Vector Algebra

Point $P$ is equidistant from the lines $A B$ and $A C$. Geometrically, this means point $P$ must lie on the angle bisector of the angle formed by vectors $A B$ and $A C$.

First, we find the unit vectors along $A B$ and $A C$ :

• $|A B|=\sqrt{3^2+1^2+(-1)^2}=\sqrt{11}$

• $|A C|=\sqrt{1^2+(-1)^2+3^2}=\sqrt{11}$

Since the magnitudes are equal, the angle bisector $v$ is simply the sum of the two vectors:

$ \begin{array}{r} v=A B+A C=(3+1) \hat{i}+(1-1) \hat{j}+(-1+3) \hat{k}=4 \hat{i}+2 \hat{k} \end{array} $

The vector $A P$ will be in the direction of the unit vector $\hat{v}$ :

$ \hat{v}=\frac{4 \hat{i}+2 \hat{k}}{\sqrt{4^2+2^2}}=\frac{4 \hat{i}+2 \hat{k}}{\sqrt{20}}=\frac{4 \hat{i}+2 \hat{k}}{2 \sqrt{5}}=\frac{2 \hat{i}+\hat{k}}{\sqrt{5}} $

Determine Vector $A P$

Given that $|A P|=\frac{\sqrt{5}}{2}$, we can write:

$ A P=|A P| \hat{v}=\frac{\sqrt{5}}{2}\left(\frac{2 \hat{i}+\hat{k}}{\sqrt{5}}\right)=\frac{1}{2}(2 \hat{i}+\hat{k})=\hat{i}+\frac{1}{2} \hat{k} $

3. Calculate the Area of Triangle $A B P$

The area of a triangle formed by two vectors $A B$ and $A P$ is given by the formula:

$ \text { Area }=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A P}| $

First, compute the cross product $A B \times A P$ :

$ \begin{gathered} \overrightarrow{A B} \times \overrightarrow{A P}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -1 \\ 1 & 0 & \frac{1}{2} \end{array}\right| \\ =\hat{i}\left(\frac{1}{2}-0\right)-\hat{j}\left(\frac{3}{2}-(-1)\right)+\hat{k}(0-1) \\ =\frac{1}{2} \hat{i}-\frac{5}{2} \hat{j}-\hat{k} \\ |\overrightarrow{A B} \times \overrightarrow{A P}|=\sqrt{\left(\frac{1}{2}\right)^2+\left(-\frac{5}{2}\right)^2+(-1)^2}=\sqrt{\frac{1}{4}+\frac{25}{4}+1}=\sqrt{\frac{26}{4}+\frac{4}{4}}=\sqrt{\frac{30}{4}}=\frac{\sqrt{30}}{2} \end{gathered} $

Finally, the area of the triangle is:

Area $=\frac{1}{2}\left(\frac{\sqrt{30}}{2}\right)=\frac{\sqrt{30}}{4} $

Given, $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$

$|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9$

$\Rightarrow $ $|\vec{a}|^2+|\vec{b}|^2-2 \vec{a} \cdot \vec{b}+|\vec{b}|^2+|\vec{c}|^2-2 \vec{b} \cdot \vec{c}+|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a}=9$

$\Rightarrow $ $6+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=9 \quad\{|\vec{a}|=|\vec{b}|=|\vec{c}|=1\}$

$ \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-\frac{3}{2} $

also,

$ |\vec{a}+\vec{b}+\vec{c}|^2=3+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) $

$=3+2\left(-\frac{3}{2}\right)=3-3=0$

So

$ \vec{a}+\vec{b}+\vec{c}=0 $

$\Rightarrow $ $\vec{a}+\vec{b}=-\vec{c}$

$\Rightarrow $ $|\vec{a}+\vec{b}|^2=|\vec{c}|^2=1$

$\Rightarrow $ $|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}=1$

$\Rightarrow $ $1+1+2 \vec{a} \cdot \vec{b}=1$

$\Rightarrow $ $ \vec{a} \cdot \vec{b}=-\frac{1}{2} $

Similarly,

$ \vec{b} \cdot \vec{c}=-\frac{1}{2} \quad \text { and } \quad \vec{a} \cdot \vec{c}=-\frac{1}{2} $

so,

$ \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=-\frac{1}{2} $

It is given that

$ |2 \vec{a}+k \vec{b}+k \vec{c}|=3 $

$\Rightarrow $ $|2 \vec{a}+k \vec{b}+k \vec{c}|^2=9$

$\Rightarrow $ $4|\vec{a}|^2+k^2|\vec{b}|^2+k^2|\vec{c}|^2+4 k \vec{a} \cdot \vec{b}+2 k^2 \vec{b} \cdot \vec{c}+4 k \vec{a} \cdot \vec{c}=9$

$\Rightarrow $ $4(1)^2+k^2(1)^2+k^2(1)^2+4 k\left(-\frac{1}{2}\right)+2 k^2\left(-\frac{1}{2}\right)+4 k\left(-\frac{1}{2}\right)=9$

$\Rightarrow $ $4+2 k^2-2 k-k^2-2 k=9$

$\Rightarrow $ $k^2-4 k-5=0$

$\Rightarrow $ $(k-5)(k+1)=0$

$\Rightarrow $ $k=5,-1$

Given

$ a=2 \hat{i}-\hat{j}-\hat{k}, b=\hat{i}+3 \hat{j}-\hat{k}, \text { and } c=2 \hat{i}+\hat{j}+3 \hat{k} . $

$v$ is in the plane of vectors $a$ and $b$,

$ \text { so } v=a+\lambda b=(2+\lambda) \hat{i}+(-1+3 \lambda) \hat{j}+(-1-\lambda) \hat{k} \text {. } $

It is given that length of projection of $v$ on $c$ is $\frac{1}{\sqrt{14}}$.

$ \begin{aligned} & \left|\frac{v \cdot c}{|c|}\right|=\frac{1}{\sqrt{14}} \\ & |v \cdot c|=\frac{1}{\sqrt{14}} \times|c| \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \quad=\frac{1}{\sqrt{14}} \times \sqrt{(2)^2+(1)^2+(3)^2} \\ & \quad=\frac{1}{\sqrt{14}} \times \sqrt{14}=1 . \\ & |((2+\lambda) \hat{i}+(-1+3 \lambda) \hat{j}+(-1-\lambda) \hat{k}) \cdot(2 \hat{i}+\hat{j}+3 \hat{k})|=1, \\ & |4+2 \lambda-1+3 \lambda-3-3 \lambda|=1, \\ & |2 \lambda|=1 \Longrightarrow \lambda= \pm \frac{1}{2} \Longrightarrow \lambda^2=\frac{1}{4} . \\ & |v|^2=|a+\lambda b|^2, \\ & |v|^2=|a|^2+\lambda^2|b|^2+2 \lambda(a \cdot b) .\{a \cdot b=0\} \\ & =6+11 \lambda^2+0 . \\ & |v|=\sqrt{6+11 \lambda^2} . \end{aligned}\\ &\text { Put value } \lambda^2=\frac{1}{4} \text { : }\\ &\begin{aligned} & |v|=\sqrt{6+\frac{11}{4}}, \\ & |v|=\sqrt{\frac{24+11}{4}}, \\ & |v|=\sqrt{\frac{35}{4}} \\ & |v|=\frac{\sqrt{35}}{2} \end{aligned} \end{aligned} $

Given

$ \begin{gathered} \vec{a}=2 \hat{i}-5 \hat{j}+5 \hat{k}, \quad \vec{b}=\hat{i}-\hat{j}+3 \hat{k} \\ a(\vec{a} \times \vec{c})+3(\vec{b} \times \vec{c})=0 \\ 2 \vec{a} \times \vec{c}+3 \vec{b} \times \vec{c}=0 \\ (2 \vec{a}+3 \vec{b}) \times \vec{c}=0 \end{gathered} $

This means $2 \vec{a}+3 \vec{b}$ and $\vec{c}$ are parallel.

$ \Rightarrow \vec{c}=\lambda(2 \vec{a}+3 \vec{b}) $

$ \vec{c}=\lambda[2(2 \hat{i}-5 \hat{j}+5 \hat{k})+3(\hat{i}-\hat{j}+3 \hat{k})] $

$ \vec{c}=\lambda(7 \hat{i}-13 \hat{j}+19 \hat{k}) $

Given

$ (\vec{a}-\vec{b}) \cdot \vec{c}=-97 $

$ \begin{gathered} \vec{a}-\vec{b}=(2 \hat{i}-5 \hat{j}+5 \hat{k})-(\hat{i}-\hat{j}+3 \hat{k}) \\ =\hat{i}-4 \hat{j}+2 \hat{k} \end{gathered} $

Put in equation:

$ \begin{gathered} (\hat{i}-4 \hat{j}+2 \hat{k}) \cdot \lambda(7 \hat{i}-13 \hat{j}+19 \hat{k})=-97 \\ \lambda(7+52+38)=-97 \\ \lambda(97)=-97 \Rightarrow \lambda=-1 \\ \vec{c}=-1(7 \hat{i}-13 \hat{j}+19 \hat{k}) \\ \vec{c}=-7 \hat{i}+13 \hat{j}-19 \hat{k} \\ \vec{c} \times \hat{k}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -7 & 13 & -19 \\ 0 & 0 & 1 \end{array}\right| \\ =\hat{i}(13)-\hat{j}(-7)+\hat{k}(0) \\ =13 \hat{i}+7 \hat{j} \\ |\vec{c} \times \hat{k}|^2=13^2+7^2=169+49=218 \end{gathered} $

$ \begin{aligned} &\begin{aligned} & \vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=\hat{i}-\hat{j} \\ & |\vec{a}|=\sqrt{4+1+4}=3,|\vec{b}|=\sqrt{1+1}=\sqrt{2} \end{aligned}\\ &\text { It is given that } \vec{c}=\vec{a} \times \vec{b}\\ &\begin{gathered} \vec{c}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & -1 & 0 \end{array}\right| \\ \vec{c}=\hat{i}(0+2)-\hat{j}(0+2)+\hat{k}(-2-1) \\ \vec{c}=2 \hat{i}-2 \hat{j}-3 \hat{k} \end{gathered} \end{aligned} $

$ \begin{aligned} &|\vec{c}|=\sqrt{4+4+1}=\sqrt{9}=3\\ &\text { It is given that }|\vec{c} \times \vec{d}|=3 \& \text { angle between } \vec{c} \& \vec{d} \text { is } \pi / 4\\ &\begin{gathered} |\vec{c}||\vec{d}| \sin \frac{\pi}{4}=3 \\ 3|\vec{d}| \times \frac{1}{\sqrt{2}}=3 \Rightarrow|\vec{d}|=\sqrt{2} \\ |\vec{d}-\vec{a}|=\sqrt{11}(\text { given }) \\ |\vec{d}-\vec{a}|^2=11 \\ |\vec{d}|^2+|\vec{a}|^2-2 \vec{d} \cdot \vec{a}=11 \\ 2+9-2 \vec{d} \cdot \vec{a}=11 \text { use }|\vec{a}|=3 \\ 11-2 \vec{d} \cdot \vec{a}=11 \\ -2 \vec{d} \cdot \vec{a}=0 \\ \vec{d} \cdot \vec{a}=0 \end{gathered} \end{aligned} $

$ \begin{aligned} L_1 & : \vec{r}=\hat{i}+\hat{j}+3 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k}) \\ L_2 & : \vec{r}=(4 \hat{i}+\hat{j})+\mu(5 \hat{i}+2 \hat{j}+\hat{k}) \end{aligned} $

at the intersection point R , position vectors are equal.

$ (\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})=(4 \hat{i}+\hat{j})+\mu(5 \hat{i}+2 \hat{j}+\hat{k}) $

equating components of $\hat{i}, \hat{j}$ and $\hat{k}$.

$ \begin{aligned} & 1+2 \lambda=4+5 \mu \Longrightarrow 2 \lambda-5 \mu=3 \text {...... (1) } \\ & 2+3 \lambda=1+2 \mu \Longrightarrow 3 \lambda-2 \mu=-1 \text {.....(2) } \\ & 3+4 \lambda=\mu \Longrightarrow \mu-4 \lambda=3 ...... (3)\\ & \text { eqn (1) } \times 3 \text { - eqn (2) } \times 2 \\ & -15 \mu+4 \mu=9+2 \Longrightarrow-11 \mu=11 \Longrightarrow \mu=-1 \end{aligned} $

substitute in equation (1)

$ 2 \lambda-5(-1)=3 \Longrightarrow \lambda=\frac{3-5}{2}=-1 $

Now, checking equation (3) $(-1)-4(-1)=3=$ R.H.S consistent.

Now finding point R using $\lambda=-1$ in $L_1$

$ \begin{gathered} \overrightarrow{O R}=(\hat{i}+2 \hat{j}+3 \hat{k})-(2 \hat{i}+3 \hat{j}+4 \hat{k}) \\ \overrightarrow{O R}=-\hat{i}-\hat{i}-\hat{k} \end{gathered} $

Point P lies on $L_1$ so its position vector is $\quad \overrightarrow{O P}=(1+2 \lambda) \hat{i}+(2+3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}$ It is given that

$ \begin{gathered} |\overrightarrow{P R}|=\sqrt{29} \Rightarrow|\overrightarrow{P R}|^2=29 \\ \overrightarrow{P R}=\overrightarrow{O R}-\overrightarrow{O P} \\ \overrightarrow{P R}=(-2-2 \lambda) \hat{i}+(-3-3 \lambda) \hat{j}+(-4-4 \lambda) \hat{k} \\ |\overrightarrow{P R}|^2=4(1+\lambda)^2+9(1+\lambda)^2+16(1+\lambda)^2=29 \\ 29(1+\lambda)^2=29 \Rightarrow(1+\lambda)^2=1 \Rightarrow \lambda=0,-2 \end{gathered} $

point $P$ lies in first octant, means position vector of $P$ must be positive, so $\lambda=0$

$ \overrightarrow{O P}=\hat{i}+2 \hat{j}+3 \hat{k} $

similarly Q lies on $L_2$, so position vector of Q is

$ \begin{gathered} \overrightarrow{O Q}=(4+5 \mu) \hat{i}+(1+2 \mu) \hat{j}+\mu \hat{k} \\ |\overrightarrow{P Q}|=\sqrt{\frac{47}{3}} \Rightarrow|\overrightarrow{P Q}|^2=\frac{47}{3} \quad \text { (given) } \\ \overrightarrow{P Q}=\overrightarrow{O Q}-\overrightarrow{O P} \\ \overrightarrow{P Q}=(3+5 \mu) \hat{i}+(-1+2 \mu) \hat{j}+(\mu-3) \hat{k} \\ (3+5 \mu)^2+(-1+2 \mu)^2+(\mu-3)^2=\frac{47}{3} \\ 30 \mu^2+20 \mu+19=\frac{47}{3} \\ 90 \mu^2+60 \mu+57=47 \\ 90 \mu^2+60 \mu+10=0 \\ 9 \mu^2+6 \mu+1=0 \\ (3 \mu+1)^2=0 \Rightarrow \mu=-\frac{1}{3} \end{gathered} $

finding position vector of Q using $\mu=-1 / 3$

$ \begin{gathered} \overrightarrow{O Q}=\left(4-\frac{5}{3}\right) \hat{i}+\left(1-\frac{2}{3}\right) \hat{j}-\frac{1}{3} \hat{k} \\ \overrightarrow{O Q}=\frac{7}{3} \hat{i}+\frac{1}{3} \hat{j}-\frac{1}{3} \hat{k} \end{gathered} $

$ \begin{aligned} & \text { calculate } 27|Q R|^2 \\ & \qquad \overrightarrow{Q R}=\overrightarrow{O R}-\overrightarrow{O Q} \\ & \overrightarrow{Q R}=-\frac{10}{3} \hat{i}-\frac{4}{3} \hat{j}-\frac{2}{3} \hat{k} \\ & |\overrightarrow{Q R}|^2=\frac{9}{100}+\frac{9}{16}+\frac{9}{4}=\frac{9}{120} \\ & 27|\overrightarrow{Q R}|^2=\frac{9}{120} \times 27=360\end{aligned} $

Given that

$ \begin{aligned} & \vec{a} \times \vec{b}=2(\vec{a} \times \vec{c}) \\ & \vec{a} \times \vec{b}-2(\vec{a} \times \vec{c})=0 \\ & \vec{a} \times(\vec{b}-2 \vec{c})=0 \end{aligned} $

This means $\vec{a}$ and $\vec{b}-2 \vec{c}$ are parallel vectors

$ \begin{aligned} & \vec{b}-2 \vec{c}=\lambda \vec{a} \\ & \Rightarrow \vec{b}=\lambda \vec{a}+2 \vec{c}-(1) \end{aligned} $

It is given that $|\vec{a}|=1,|b|=4$ and $|\vec{c}|=2$, angle between $\vec{b}$ and $\vec{c}, \theta=60^{\circ}$.

$ |b|^2=|\lambda \vec{a}+2 \vec{c}|^2 $

$\Rightarrow $ $|b|^2=\lambda^2|a|^2+4|c|^2+4 \lambda(\vec{a} \cdot \vec{c})$

$\Rightarrow $ $16=\lambda^2(1)^2+4(2)^2+4 \lambda(\vec{a} \cdot \vec{c})$

$\Rightarrow $ $\lambda^2+4 \lambda(\vec{a} \cdot \vec{c})=0$ .......(2)

Taking dot product with $\overrightarrow{\mathrm{c}}$ in equation (1)

$ \vec{b} \cdot \vec{c}=(\lambda \vec{a}+2 \vec{c}) \cdot \vec{c} $

$\Rightarrow $ $|\vec{b}||\vec{c}| \cos 60^{\circ}=\lambda(\vec{a} \cdot \vec{c})+2|\vec{c}|^2$

$\Rightarrow $ $4 \times 2 \times 1 / 2=\lambda(\vec{a} \cdot \vec{c})+2(2)^2$

$\Rightarrow $ $4=\lambda(\vec{a} \cdot \vec{c})+8$

$\Rightarrow $ $ \lambda(\vec{a} \cdot \vec{c})=-4 $

From (2) and (3)

$\lambda^2+4(-4)=0$

$\Rightarrow $ $\lambda^2-16=0$

$\Rightarrow $ $ \lambda= \pm 4 $

from (3)

$ \vec{a} \cdot \vec{c}=-4 / \lambda $

if $\lambda=4$, then $\vec{a} \cdot \vec{c}=-1$

if $\lambda=-4$, then $\vec{a} \cdot \vec{c}=1$

$ |\vec{a} \cdot c|=1 $

$ \begin{aligned} &\text { Given }\\ &\begin{gathered} \vec{a}=\hat{i}-2 \hat{j}+3 \hat{k} \\ \vec{b}=2 \hat{i}+\hat{j}-\hat{k} \\ \vec{c}=\lambda \hat{i}+\hat{j}+\hat{k} \\ \vec{v}=\vec{a} \times \vec{b} \\ \vec{v}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{array}\right| \\ \vec{v}=\hat{i}(2-3)-\hat{j}(-1-6)+\hat{k}(1+4) \\ \vec{v}=-\hat{i}+7 \hat{j}+5 \hat{k} \\ \vec{v} \cdot \vec{c}=11 \end{gathered} \end{aligned} $

$ \begin{gathered} (-\hat{i}+7 \hat{j}+5 \hat{k}) \cdot(\lambda \hat{i}+\hat{j}+\hat{k})=11 \\ -\lambda+7+5=11 \\ -\lambda+12=11 \\ \lambda=1 \\ \vec{c}=\hat{i}+\hat{j}+\hat{k} \\ p=\frac{|\vec{b} \cdot \vec{c}|}{|\vec{c}|} \\ \vec{b} \cdot \vec{c}=(2 \hat{i}+\hat{j}-\hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k}) \\ =2+1-1=2 \\ |\vec{c}|=\sqrt{1^2+1^2+1^2}=\sqrt{3} \\ p=\frac{2}{\sqrt{3}} \\ p^2=\frac{4}{3} \\ 9 p^2=12 \end{gathered} $

$ \begin{aligned} &\begin{aligned} & (\vec{a}-\vec{b}) \cdot \vec{d}=\vec{a} \cdot \vec{d}-\vec{b} \cdot \vec{d} \\ & \because \vec{d}=\vec{c} \times \vec{a} \Rightarrow \vec{a} \cdot \vec{d}=[\vec{a} \vec{c} \vec{a}]=0 \\ & \text { And } \vec{b} \cdot \vec{d}=\vec{b} \cdot(\vec{c} \times \vec{a}) \\ & =\vec{b} \cdot((\vec{a} \times \vec{b}) \times \vec{a}) \\ & \Rightarrow(\vec{a}-\vec{b}) \cdot \vec{d}=\vec{b} \cdot(\vec{a} \times(\vec{a} \times \vec{b})) \\ & =\vec{b} \cdot[\vec{a} \cdot \vec{b} \vec{a}-\vec{a} \cdot \vec{a} \vec{b}] \\ & =(\vec{a} \cdot \vec{b})^2-(\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{b}) \\ & =(-8)^2-6 \times 11=64-66=-2 \end{aligned}\\ &\text { Option (2) is correct. } \end{aligned} $

$ \begin{aligned} & \bar{c}=\bar{a} \times \bar{b}=\left|\begin{array}{ccc} \bar{i} & \bar{j} & \bar{k} \\ 2 & 1 & 1 \\ 0 & \lambda & 2 \end{array}\right|=(2-\lambda) \bar{i}-\bar{j}(4)+(2 \lambda) \bar{k} \\ & |\bar{c}|^2=53 \Rightarrow(2-\lambda)^2+16+(2 \lambda)^2=53=5 \lambda^2-4 \lambda-33=0 \\ & \Rightarrow 5 \lambda^2-15 \lambda+11 \lambda-33=0 \\ & \Rightarrow 5 \lambda(\lambda-3)+11(\lambda-3)=0 \\ & \Rightarrow(\lambda-3)(5 \lambda+11)=0 \\ & \Rightarrow \lambda=3, \lambda \neq-\frac{11}{5} \,\,\,\,\,(\because \lambda \in Z)\\ & \therefore \bar{c}=(-1,-4,6) \end{aligned} $

Let $\therefore \bar{d}=(0, y, z)$

Given $|\bar{d}|=2 \Rightarrow y^2+z^2=4$

$ \begin{aligned} & \bar{c} . \bar{d}=0-4 y+6 z \\ & (\bar{c} . \bar{d})^2=(-4 y+6 z)^2 \\ & =16 y^2+36 z^2-48 y z \\ & =16 y^2+36\left(4-y^2\right)-48 y \cdot \sqrt{4-y^2} \end{aligned} $

Let $\mathrm{f}(\mathrm{y})=-20 y^2+144-48 y \sqrt{4-y^2}$

$ \begin{aligned} & f^1(y)=-40 y-48\left(\sqrt{4-y^2}\right)-48 y\left(\frac{1}{2 \sqrt{4-y^2}}\right)(-2 y) \\ & =-40 y-48\left(\frac{4-y^2-y^2}{\sqrt{4-y^2}}\right) \\ & f^1(y)=0 \Leftarrow 40 y=96\left(\frac{y^2-2}{\sqrt{4-y^2}}\right) \\ & \Rightarrow 5 y \sqrt{4-y^2}=12\left(y^2-2\right) \\ & \Rightarrow 25 y^2\left(4-y^2\right)=144\left(y^4+4-4 y^2\right) \\ & =169 y^4-676 y^2+576=0 \\ & =169 y^4-208 y^2-468 y^2+576=0 \\ & \Rightarrow 13 y^2\left(13 y^2-16\right)-36\left(13 y^2-16\right)=0 \\ & \Rightarrow y^2=\frac{16}{13}(\text { or }) \frac{36}{13} \\ & \Rightarrow y= \pm \frac{4}{\sqrt{13}}(\text { or }) \pm \frac{6}{\sqrt{13}} \\ & \therefore \text { Maximum of }(\bar{c} . \bar{d})^2=208 \end{aligned} $

$ \begin{aligned} & \overline{A B}=2 \bar{i}+4 \bar{j}-5 \bar{k} \\ & \overline{A D}=\bar{i}+2 \bar{j}+\lambda \bar{k} \\ & \bar{V}=\bar{i}+\bar{j}+\bar{k} \\ & \bar{V} \cdot \overline{A C} \\ & \frac{|\overline{A C}|}{} \end{aligned} $

$ \begin{aligned} & \overline{A C}=\overline{A B}+\overline{B C} \\ & 18 \lambda=54 \\ & 18 \lambda=54 \end{aligned} $

$ \begin{aligned} & \Rightarrow \vec{r}=\vec{q}+\vec{p} \\ & |\vec{r}|=\sqrt{q^2+p^2+2 p q \cos \theta} \\ & |r|^2=4+12+2 \times \sqrt{3} \times 2 \times \frac{1}{\sqrt{3}} \\ & |r|^2=24 \\ & \left|\vec{p} \times\left(\vec{q}-3(\vec{q}-\vec{p})^2\right)\right|^2+3|\vec{r}|^2=|-2 \vec{p} \times \vec{q}|^2+3|\vec{r}|^2 \\ & =4 \times|\vec{p}||\vec{q}|^2 \sin ^2 \theta+3 \times 24=4 \times 12 \times 4 \times\left(\frac{2}{3}\right)+72=128+72=200 \end{aligned} $

$ \begin{aligned} &\text { Let } \vec{c}=x \hat{i}+y \hat{j}+z \hat{k}\\ &\begin{aligned} & \vec{a} \times \vec{c}=\vec{b}\left|\begin{array}{ccc} \hat{i} & \hat{j} & k \\ -1 & 2 & 2 \\ x & y & z \end{array}\right| \\ & =(2 z-2 y) \hat{i}+(z+2 x) \hat{j}+(-y-2 x) \hat{k} \\ & \equiv 8 \hat{i}+7 \hat{j}-3 \hat{k} \end{aligned} \end{aligned} $

$ \begin{aligned} \Rightarrow & 2 z-2 y=8 \\ & \left.\begin{array}{l} z+2 x=7 \\ -y-2 x=3 \end{array}\right\} y=3-2 x, z=7-2 x \end{aligned} $

$ \begin{aligned} & \vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=4 \\ & x+y+z=4 \\ & \text { Now } x+3-2 x+7-2 x=4 \\ & \Rightarrow \quad x=2 \\ & \quad y=-1 \\ & \quad z=3 \\ & \Rightarrow \quad \vec{c}=2 \hat{i}-\hat{j}+3 \hat{k} \\ & |\vec{a}+\vec{c}|^2=|\hat{i}+\hat{j}+5 \hat{k}|^2 \\ & =27 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}=\lambda \\ & (2 \lambda-1,3 \lambda+3,1-\lambda) \\ & \overrightarrow{P F}=(2 \lambda-6) \hat{i}+(3 \lambda-1) \hat{j}+(-\lambda-1) \hat{k} \\ & \overrightarrow{P F} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})=0 \\ & 4 \lambda-12+9 \lambda-3+\lambda+1=0 \Rightarrow \lambda=1 \\ & \Rightarrow F \equiv(1,6,0) \\ & \Rightarrow \text { length of projection of } \hat{i}+6 \hat{j} \text { on } 6 \hat{i}+2 \hat{j}+3 \hat{k} \\ & =\left|\frac{(\hat{i}+6 \hat{j}) \cdot(6 \hat{i}+2 \hat{j}+3 \hat{k})}{|6 \hat{i}+2 \hat{j}+3 \hat{k}|}\right| \\ & =\frac{1}{7}(6+12)=\frac{18}{7} \end{aligned}\\ &\text { Option (3) is correct. } \end{aligned} $

$ \begin{aligned} & |\vec{c}+\vec{d}|=\sqrt{29} \\ & \vec{c} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})-(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{d}=0 \end{aligned} $

$ \begin{aligned} & (\vec{c}+\vec{d}) \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=0 \\ & \Rightarrow \vec{c}+\vec{d}=\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k}) \\ & \Rightarrow \sqrt{29}=|\lambda| \sqrt{29} \\ & \Rightarrow \lambda= \pm 1 \Rightarrow \vec{c}+\vec{d}= \pm(2 \hat{i}+3 \hat{j}+4 \hat{k}) \\ & \Rightarrow(\vec{c}+\vec{d}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})= \pm(-14+6+12)= \pm 4 \\ & \Rightarrow \lambda_1=4, \lambda_2=-4 \end{aligned} $

Given equation represents a circle

$ \begin{aligned} & \Rightarrow K^2=3 K+\frac{\lambda_2}{2} \text { and } K^2-5 K+\lambda_1=0 \\ & \Rightarrow K^2=3 K-2 \text { and } K^2-5 K+4=0 \\ & \Rightarrow K=1,2 \text { and } K=4,1 \\ & \Rightarrow K=1 \end{aligned} $

⇒ option (4) is correct.

Let us find the area of triangle $OPR$ step by step.

Given:

• Position vector of $P$ is $\vec{a}$

• Position vector of $Q$ is $\vec{b}$

• Position vector of $R$ is $\vec{a}+\vec{b}$

So, for triangle $OPR$, the two side vectors from $O$ are:

$ \overrightarrow{OP}=\vec{a}, \qquad \overrightarrow{OR}=\vec{a}+\vec{b} $

Hence, area of triangle $OPR$ is

$ \frac{1}{2}\left|\vec{a}\times(\vec{a}+\vec{b})\right| $

Now,

$ \vec{a}\times(\vec{a}+\vec{b})=\vec{a}\times\vec{a}+\vec{a}\times\vec{b}=\vec{a}\times\vec{b} $

since $\vec{a}\times\vec{a}=0$.

Therefore,

$ \text{Area of } \triangle OPR=\frac{1}{2}|\vec{a}\times\vec{b}| $

So we only need to find $|\vec{a}\times\vec{b}|$.

We are given:

$ |\vec{a}+\vec{b}|=\sqrt{21} $

Squaring both sides,

$ |\vec{a}+\vec{b}|^2=21 $

So,

$ |\vec{a}|^2+|\vec{b}|^2+2\vec{a}\cdot\vec{b}=21 \qquad (1) $

Also,

$ |\vec{a}-\vec{b}|=3 $

Squaring,

$ |\vec{a}-\vec{b}|^2=9 $

Thus,

$ |\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b}=9 \qquad (2) $

Now subtract $(2)$ from $(1)$:

$ 4\vec{a}\cdot\vec{b}=12 $

So,

$ \vec{a}\cdot\vec{b}=3 $

Now use the condition that $\vec{a}$ is perpendicular to $(\vec{a}-\vec{b})$.

Therefore,

$ \vec{a}\cdot(\vec{a}-\vec{b})=0 $

$ |\vec{a}|^2-\vec{a}\cdot\vec{b}=0 $

Since $\vec{a}\cdot\vec{b}=3$, we get

$ |\vec{a}|^2=3 $

So,

$ |\vec{a}|=\sqrt{3} $

Now use equation $(2)$:

$ |\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b}=9 $

Substitute $|\vec{a}|^2=3$ and $\vec{a}\cdot\vec{b}=3$:

$ 3+|\vec{b}|^2-6=9 $

$ |\vec{b}|^2=12 $

Now use the identity

$ |\vec{a}\times\vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2 $

Substitute the values:

$ |\vec{a}\times\vec{b}|^2=(3)(12)-3^2=36-9=27 $

Therefore,

$ |\vec{a}\times\vec{b}|=\sqrt{27}=3\sqrt{3} $

Hence area of triangle $OPR$ is

$ \frac{1}{2}|\vec{a}\times\vec{b}|=\frac{1}{2}(3\sqrt{3}) $

$ \boxed{\frac{3\sqrt{3}}{2}} $

So, the correct option is:

$ \boxed{\text{Option C}} $

Given the matrix

$ M = \begin{bmatrix} \alpha & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \beta & \frac{1}{\sqrt{3}} \\ \gamma & \delta & \mu \end{bmatrix}, $

and $ MM^{T} = I $, this means that the rows of $ M $ are orthonormal vectors (that is, each has unit length and they are mutually perpendicular).

Let

$ \vec{u} = \alpha \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \gamma \hat{k}, \quad \vec{v} = \frac{1}{\sqrt{2}} \hat{i} + \beta \hat{j} + \delta \hat{k}, \quad \vec{w} = -\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \mu \hat{k}. $

Length condition for each row (unit vectors)

For $ \vec{u} $, the condition $ |\vec{u}| = 1 $ gives

$ \alpha^2 + \frac{1}{3} + \gamma^2 = 1 $

Thus,

$ \alpha^2 + \gamma^2 = \frac{2}{3} \tag{1} $

For $ \vec{v} $, the condition $ |\vec{v}| = 1 $ gives

$ \frac{1}{2} + \beta^2 + \delta^2 = 1 $

Therefore,

$ \beta^2 + \delta^2 = \frac{1}{2} \tag{2} $

For $ \vec{w} $, the condition $ |\vec{w}| = 1 $ gives

$ \frac{1}{2} + \frac{1}{3} + \mu^2 = 1 $

Hence,

$ \mu^2 = \frac{1}{6} \tag{3} $

Using the third row’s condition

The third row of $ M $ also has unit length, so

$ \gamma^2 + \delta^2 + \mu^2 = 1 $

Substitute $ \mu^2 = \frac{1}{6} $:

$ \gamma^2 + \delta^2 = 1 - \frac{1}{6} = \frac{5}{6} $

So, for (P), the value of $ \gamma^2 + \delta^2 = \frac{5}{6}. $

Finding $ x $ when $ x\vec{u} + y\vec{v} + z\vec{w} = \hat{j} $

Taking the dot product with $ \vec{u} $ on both sides:

$ x(\vec{u} \cdot \vec{u}) + y(\vec{v} \cdot \vec{u}) + z(\vec{w} \cdot \vec{u}) = \vec{u} \cdot \hat{j} $

Since the vectors are mutually perpendicular, $ \vec{u} \cdot \vec{v} = 0 $ and $ \vec{u} \cdot \vec{w} = 0 $. So,

$ x(1) = \frac{1}{\sqrt{3}} $

Therefore, $ x = \frac{1}{\sqrt{3}}. $

Value of $ |\vec{u} \cdot (\vec{v} \times \vec{w})| $

For an orthogonal matrix, the determinant is $ \pm 1 $, meaning the scalar triple product equals 1 in magnitude. Hence,

$ |\vec{u} \cdot (\vec{v} \times \vec{w})| = 1. $

Value of $ |\vec{u} \times (\vec{v} \times \vec{w})| $

Using vector identity $ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} $, and since the vectors are perpendicular, both $ \vec{u} \cdot \vec{v} = 0 $ and $ \vec{u} \cdot \vec{w} = 0 $. Thus,

$ \vec{u} \times (\vec{v} \times \vec{w}) = 0, $

and its magnitude is 0.

Final Matching:

(P) → (5) $ \dfrac{5}{6} $
(Q) → (4) $ \dfrac{1}{\sqrt{3}} $
(R) → (2) $ 1 $
(S) → (1) $ 0 $

Let vector $\vec{p}$ in plane of $\vec{a} \& \vec{b}=K(\vec{a}+\lambda \vec{b})$

$\begin{aligned} & \overrightarrow{\mathrm{p}} \perp \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{a}}=0 \\ & \Rightarrow \mathrm{~K}(\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{a}}=0 \\ & \Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}=0 \\ & \Rightarrow 6+\lambda(3)=0 \\ & \Rightarrow \lambda=-2 \\ & \Rightarrow \overrightarrow{\mathrm{p}}=(-3 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}) \end{aligned}$

Unit vector $\rightarrow \pm \frac{(-\hat{\mathrm{i}}+\hat{\mathrm{k}})}{\sqrt{2}}$

$\frac{|\bar{a}+\bar{b}|+|\bar{a}-\bar{b}|}{|\bar{a}+\bar{b}|-|\bar{a}-\bar{b}|}=\sqrt{2}+1$

Apply componendo and dividendo

$\begin{aligned} & \Rightarrow \frac{2|\bar{a}+\bar{b}|}{2|\bar{a}-\bar{b}|}=\frac{\sqrt{2}+2}{\sqrt{2}} \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}|=(1+\sqrt{2})|\overline{\mathrm{a}}-\overline{\mathrm{b}}| \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}|^2=(3+2 \sqrt{2})|\overline{\mathrm{a}}-\overline{\mathrm{b}}|^2 \\ & \Rightarrow 2|\overline{\mathrm{a}}|^2+2 \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}=(3+2 \sqrt{2})\left(2|\overline{\mathrm{a}}|^2-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}\right) \\ & \Rightarrow 2|\overline{\mathrm{a}}|^2(2+2 \sqrt{2})=2 \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}(4+2 \sqrt{2}) \\ & \Rightarrow \frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}}{|\overline{\mathrm{a}}|^2}=\frac{2+2 \sqrt{2}}{4+2 \sqrt{2}}=\frac{1}{\sqrt{2}} \end{aligned}$

Now

$\begin{aligned} & \frac{|\overline{\mathrm{a}}+\overline{\mathrm{b}}|^2}{|\overline{\mathrm{a}}|^2}=1+\frac{|\overline{\mathrm{b}}|^2}{|\overline{\mathrm{a}}|^2}+\frac{2 \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}}{|\overline{\mathrm{a}}|^2} \\ & =1+1+2\left(\frac{1}{\sqrt{2}}\right)=2+\sqrt{2} \end{aligned}$

To solve the problem, we begin by examining the vector $\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})$.

Let's calculate:

$\vec{c} \cdot \hat{a} = 3 + 6(\hat{a} \cdot \hat{b})$.

$\vec{c} \cdot \hat{b} = 3(\hat{a} \cdot \hat{b}) + 6$.

We need to find the value of $9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$.

Substituting the expressions, we get:

$ 9(\vec{c} \cdot \hat{a}) = 9(3 + 6\hat{a} \cdot \hat{b}) = 27 + 54(\hat{a} \cdot \hat{b}) $

$ 3(\vec{c} \cdot \hat{b}) = 3(3\hat{a} \cdot \hat{b} + 6) = 9\hat{a} \cdot \hat{b} + 18 $

Therefore,

$ 9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}) = (27 + 54(\hat{a} \cdot \hat{b})) - (9\hat{a} \cdot \hat{b} + 18) $

Simplifying,

$ = 27 - 18 + 54(\hat{a} \cdot \hat{b}) - 9(\hat{a} \cdot \hat{b}) $

$ = 9 + 45(\hat{a} \cdot \hat{b}) $

Given $\sin{\theta} = \frac{\sqrt{65}}{9}$, and knowing that for unit vectors $\cos{\theta} = \hat{a} \cdot \hat{b}$, we use the identity $(\cos{\theta})^2 = 1 - (\sin{\theta})^2$:

$ (\hat{a} \cdot \hat{b})^2 = 1 - \left(\frac{\sqrt{65}}{9}\right)^2 $

$ = 1 - \frac{65}{81} $

$ = \frac{16}{81} $

Thus, $\hat{a} \cdot \hat{b} = \frac{4}{9}$.

Substitute back into the equation:

$ 9 + 45 \times \frac{4}{9} = 9 + 20 = 29 $

Therefore, the value is 29.

$\begin{aligned} & \vec{u} \cdot \vec{v}=|u| \cdot|v| \cdot \cos \theta \\ & \Rightarrow 6-1=\sqrt{10} \cdot \sqrt{5+\lambda^2} \cdot \frac{\sqrt{5}}{2 \sqrt{7}} \\ & \Rightarrow 1=\sqrt{2} \cdot \sqrt{5+\lambda^2} \cdot \frac{1}{2 \sqrt{7}} \\ & \Rightarrow 14=5+\lambda^2 \\ & \Rightarrow \lambda^2=9 \\ & \Rightarrow \lambda=3 \\ & v_1=k \vec{u} \\ & \vec{v}=\vec{v}_1+\vec{v}_2 \\ & \Rightarrow \vec{v}=k \vec{u}+\vec{v}_2 \\ & \vec{v} \cdot \vec{u}=k \cdot|\vec{u}|^2 \end{aligned}$

$\begin{aligned} & \Rightarrow 5=k \cdot 10 \Rightarrow k=\frac{1}{2} \\ & \therefore \quad \vec{v}_1=\frac{\vec{u}}{2}=\frac{3 \hat{i}}{2}-\frac{\hat{j}}{2} \\ & \left|\vec{v}_1\right|^2=\frac{10}{4} \\ & \vec{v}_2=\vec{v}-\vec{v}_1 \\ & =\frac{1}{2} \hat{i}+\frac{3 \hat{j}}{2}-3 \hat{k} \\ & \left|\vec{v}_2\right|^2=\frac{10}{4}+9 \\ & \left|\vec{v}_1\right|^2+\left|\vec{v}_2\right|^2=\frac{10}{4}+\frac{10}{4}+9=14 \end{aligned}$

To solve the problem, we start with the given equation:

$(\vec{a} - \vec{c}) \times \vec{b} = -18 \hat{i} - 3 \hat{j} + 12 \hat{k}.$

Expanding this, we have:

$\vec{a} \times \vec{b} - \vec{c} \times \vec{b} = -18 \hat{i} - 3 \hat{j} + 12 \hat{k}.$

Let $\vec{d} = \vec{c} \times \vec{b}$. Substituting this into the equation, we get:

$\vec{a} \times \vec{b} + \vec{d} = -18 \hat{i} - 3 \hat{j} + 12 \hat{k}.$

Now, we take the dot product of both sides with $\vec{a}$:

$\vec{a} \cdot (\vec{a} \times \vec{b}) + \vec{a} \cdot \vec{d} = \vec{a} \cdot (-18 \hat{i} - 3 \hat{j} + 12 \hat{k}).$

Since the dot product $\vec{a} \cdot (\vec{a} \times \vec{b})$ is zero (the dot product of any vector with the cross product of itself with another vector is zero), we have:

$\vec{a} \cdot \vec{d} = -36 + 9 + 12 = -15.$

Thus, the magnitude of $\vec{a} \cdot \vec{d}$ is:

$|\vec{a} \cdot \vec{d}| = 15.$

Projection of $\vec{a}$ on $\vec{v}$

$\begin{aligned} & =\frac{\vec{a} \cdot \vec{v}}{|\vec{v}|}- \\ & \Rightarrow \frac{\vec{a} \cdot(2 \hat{i}-\hat{j}+2 \hat{k})}{3}=\frac{\vec{a} \cdot \hat{k}}{1}=\frac{\vec{a} \cdot(\hat{i}+2 \hat{j}-2 \hat{k})}{3} \\ & \Rightarrow \vec{a} \cdot(2 \hat{i}-\hat{j}-\hat{k})=0 \text { and } \vec{a} \cdot(\hat{i}+2 \hat{j}-5 \hat{k})=0 \\ & \Rightarrow \vec{a} \perp(2 \hat{i}-\hat{j}-\hat{k}) \text { and }(\hat{i}+2 \hat{j}-5 \hat{k}) \\ & \Rightarrow \vec{a} \|(2 \hat{i}-\hat{j}-\hat{k}) \times(\hat{i}+2 \hat{j}-5 \hat{k}) \\ & \Rightarrow \vec{a}= \pm k\left|\begin{array}{ccc} \hat{i} & -\hat{j} & \hat{k} \\ 2 & -1 & -1 \\ 1 & 2 & -5 \end{array}\right|= \pm k(7 \hat{i}+9 \hat{j}-5 \hat{k}) \\ & \Rightarrow \text { Unit vector will be } \frac{1}{\sqrt{155}}(7 \hat{i}+9 \hat{j}+5 \hat{k}) \end{aligned}$

$\begin{aligned} \text { Let } & \overrightarrow{\mathrm{v}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \begin{aligned} \overrightarrow{\mathrm{b}} & \times \overrightarrow{\mathrm{c}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{array}\right| \\ & =-7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\ & =-7(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}) \end{aligned} \end{aligned}$

Now $\hat{a}=\frac{\hat{\mathbf{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{3}}$ or $\frac{-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}$

$\qquad\qquad\downarrow \qquad \qquad\downarrow$

$\begin{aligned} \cos \theta= & \frac{\hat{\mathrm{a}} \cdot \overrightarrow{\mathrm{v}}}{|\overrightarrow{\mathrm{v}}|}=\frac{1-1-1}{\sqrt{3} \sqrt{3}}=\frac{-1}{3} \quad \cos \theta=\frac{\hat{\mathrm{a}} \cdot \overrightarrow{\mathrm{v}}}{|\overrightarrow{\mathrm{v}}|}=\frac{-1+1+1}{3}=\frac{1}{3}\quad\text{(rejected)}\\ & \Rightarrow \hat{\mathrm{a}}=\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{3}} \end{aligned}$

$\begin{aligned} & \text { Now } \cos \frac{\pi}{3}=\frac{\hat{\mathrm{a}} \cdot(\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}})}{\sqrt{1+\alpha^2+1}} \\ & \Rightarrow \frac{1}{2}=\frac{1-\alpha-1}{\sqrt{3} \sqrt{\alpha^2+2}} \\ & \Rightarrow \frac{\sqrt{3}}{2} \sqrt{\alpha^2+2}=-\alpha \quad(\therefore \alpha<0) \\ & 3 \alpha^2+6=4 \alpha^2 \\ & \Rightarrow \alpha=-\sqrt{6} \end{aligned}$

$\begin{aligned} & L_1: \overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & \Rightarrow \overrightarrow{\mathrm{r}}=(\lambda-1) \hat{\mathrm{i}}+2(\lambda+1) \hat{\mathrm{j}}+(\lambda+1) \hat{\mathrm{k}} \\ & L_2: \overrightarrow{\mathrm{r}}=(\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ & \Rightarrow \overrightarrow{\mathrm{r}}=2 \mu \hat{\mathrm{i}}+(1+7 \mu) \hat{\mathrm{j}}+(1+3 \mu) \hat{\mathrm{k}} \end{aligned}$

For point of intersection equating respective components

$\begin{aligned} &\begin{aligned} & \Rightarrow \lambda-1=2 \mu \quad\text{....... (1)}\\ & 2(\lambda+1)=1+7 \mu \quad\text{...... (2)}\\ & \lambda+1=1+3 \mu \quad\text{..... (3)} \end{aligned}\\ &\text { We get } \end{aligned}$

$\begin{aligned} & \Rightarrow \lambda=3 \text { and } \mu=1 \\ & \Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & L_3: \overrightarrow{\mathrm{r}}=2 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}+\alpha(3 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \\ & \text { For } \alpha=2, \overrightarrow{\mathrm{r}}=8 \hat{\mathrm{i}}+26 \hat{\mathrm{j}}+12 \hat{\mathrm{k}} \end{aligned}$

$\begin{aligned} & \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{~b}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}} \\ & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=0 \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow \overrightarrow{\mathrm{c}}=\lambda(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \\ & \overrightarrow{\mathrm{c}}=\lambda(5 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \ldots . .(1) \\ & |\overrightarrow{\mathrm{c}}|^2=\lambda^2(25+36+16) \\ & |\overrightarrow{\mathrm{c}}|^2=77 \lambda^2 \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}) \cdot(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=168 \\ & \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{~b}}+|\overrightarrow{\mathrm{c}}|^2=168 \\ & 14+\overrightarrow{\mathrm{c}} \cdot(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})+77 \lambda^2=168 \end{aligned}$

using equation (1)

$\begin{aligned} & \lambda|5 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}|^2+77 \lambda^2=154 \\ & 77 \lambda+77 \lambda^2-154=0 \\ & \lambda^2+\lambda-2=0 \\ & \lambda=-2,1 \end{aligned}$

$\therefore$ Maximum value of $|\overrightarrow{\mathrm{c}}|^2$ occurs when $\lambda=-2$

$\begin{aligned} & |\overrightarrow{\mathrm{c}}|^2=77 \lambda^2 \\ & =77 \times 4 \\ & =308 \end{aligned}$

$\begin{aligned} & \text { let } \\ & \overrightarrow{\mathrm{a}}_{11}=\text { component of } \overrightarrow{\mathrm{a}} \text { along } \overrightarrow{\mathrm{b}} \\ & \overrightarrow{\mathrm{a}}_1=\text { component of } \overrightarrow{\mathrm{a}} \text { perpendicular to } \overrightarrow{\mathrm{b}} \\ & \overrightarrow{\mathrm{a}}_{11}=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & \overrightarrow{\mathrm{a}}_1=\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}}) \\ & \because \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}}_{11}+\overrightarrow{\mathrm{a}}_1 \\ & \therefore \overrightarrow{\mathrm{a}}=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})+\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}}) \\ & \quad=\frac{44}{11} \hat{\mathrm{i}}+\frac{11}{11} \hat{\mathrm{j}}-\frac{33}{11} \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{a}}=4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}} \quad \\ & \begin{array}{l} \alpha=4 \qquad \beta=1\qquad \gamma=-3\\ \alpha^2+\beta^2+\gamma^2=16+1+9=26 \end{array} \end{aligned}$

$\begin{aligned} &\text { Equation of angle bisector : } \mathrm{x}-\mathrm{y}=0\\ &\begin{aligned} & \left|\frac{\mathrm{a}(1-\mathrm{a})}{\sqrt{2}}\right|=\frac{9}{\sqrt{2}} \Rightarrow \mathrm{a}=5 \text { or }-4 \\ & \text { Sum }=5+(-4)=1 \end{aligned} \end{aligned}$

We know that,

$\begin{aligned} &\begin{aligned} & \mathrm{O} \text { (orthocentre) } \frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4} \\ & \mathrm{C}(\text { circum centre) } \alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}} \\ & \mathrm{C} \text { (centroid) }=\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{3} \end{aligned}\\ &\text { by relation }\\ &\begin{aligned} & \Rightarrow 2(\alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}})+\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4}=3\left(\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{3}\right) \\ & \Rightarrow 8(\alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}})=3(\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}) \\ & \Rightarrow 8 \alpha=3,8 \beta=3,8 \gamma=3 \\ & \alpha=\frac{3}{8}, \beta=\frac{3}{8}, \gamma=\frac{3}{8} \\ & \therefore \alpha+2 \beta+3 \gamma \\ & \frac{3}{8}+\frac{6}{8}+\frac{15}{8}=\frac{24}{8}=3 \end{aligned} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{a}} \times(\hat{\mathrm{i}}-3 \hat{\mathrm{k}}) \\ & =\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & -1 & 2 \\ 1 & 0 & -2 \end{array}\right|=2 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \hat{\mathrm{k}}=8 \hat{\mathrm{i}}-2 \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{c}}-2 \hat{\mathrm{j}}=8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}} \end{aligned}\\ &\text { Projection of }(\hat{i}-2 \hat{j}) \text { on } \vec{a}\\ &\begin{aligned} & (\overrightarrow{\mathrm{c}}-2 \hat{\mathrm{j}}) \cdot \hat{\mathrm{a}}=\frac{\langle 8,-4,0\rangle \cdot\langle 3,-1,2\rangle}{\sqrt{14}} \\ & =\frac{28}{\sqrt{14}}=2 \sqrt{14} \end{aligned} \end{aligned}$

$\begin{aligned} & \overrightarrow{\mathrm{c}}=\lambda(\overrightarrow{\mathrm{b}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})) \\ & =\lambda((\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{~b}}) \overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}) \overrightarrow{\mathrm{b}}) \\ & =\lambda(11 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{~b}})=\lambda(11 \mathrm{i}+22 \mathrm{j}+33 \mathrm{k}-6 \mathrm{i}-2 \mathrm{j}+2 \mathrm{k}) \\ & =\lambda(5 \mathrm{i}+20 \mathrm{j}+35 \mathrm{k}) \\ & =5 \lambda(5 \mathrm{i}+4 \mathrm{j}+7 \mathrm{k}) \\ & =\text { Given } \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=5 \\ & =5 \lambda(1+8+21)=5=\lambda=\frac{1}{30} \\ & \Rightarrow \overrightarrow{\mathrm{c}}=\frac{1}{6}(\mathrm{i}+4 \mathrm{j}+7 \mathrm{k}) \\ & |\overrightarrow{\mathrm{c}}|=\frac{\sqrt{1+16+49}}{6}=\sqrt{\frac{11}{6}} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \mathrm{A} \equiv\left(\frac{5 \mathrm{r}-1}{\mathrm{r}+1}, \frac{5 \mathrm{r}-1}{\mathrm{r}+1}, \frac{10 \mathrm{r}+2}{\mathrm{r}+1}\right) \\ & (\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OA}})-\frac{|\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OA}}|^2}{5}=10 \quad\text{.... (1)}\\ & \overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OA}}=\frac{10}{\mathrm{r}+1}(15 \mathrm{r}+1) \\ & |\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OA}}|^2=\frac{\mathrm{r}^2}{(\mathrm{r}+1)^2}(800) \end{aligned}\\ &\text { so by equation (1) }\\ &\begin{aligned} & \frac{10}{r+1}(15 r+1)-\frac{1}{5} \frac{r^2(800)}{(r+1)^2}=10 \\ & 2 r^2-14 r=0 \\ & r=7, r \neq 0 \end{aligned} \end{aligned}$

Area of $\triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$

$=\frac{1}{2}|5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}|=\frac{1}{2} \sqrt{35}$

volume of tetrahedron

$\begin{aligned} & =\frac{1}{3} \times \text { Base area } \times \mathrm{h}=\frac{\sqrt{805}}{6 \sqrt{2}} \\ & \frac{1}{3} \times \frac{1}{2} \sqrt{35} \times \mathrm{h}=\frac{\sqrt{805}}{6 \sqrt{2}} \\ & \mathrm{~h}=\sqrt{\frac{23}{2}} \end{aligned}$

$\mathrm{AE}^2=\mathrm{AD}^2-\mathrm{DE}^2=\frac{13}{18} \therefore \mathrm{AE}=\sqrt{\frac{13}{18}}$

$\begin{aligned} & \overrightarrow{\mathrm{AE}}=|\mathrm{AE}| \cdot\left(\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{\sqrt{26}}\right) \\ & =\sqrt{\frac{13}{18}} \cdot\left(\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{\sqrt{26}}\right) \\ & =\sqrt{\frac{13}{18}} \cdot\left(\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{\sqrt{26}}\right)=\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{6} \\ & \text { P.V. of } \mathrm{E}=\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{6}+\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}=\frac{1}{6}(7 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \end{aligned}$

$\begin{aligned} & \overrightarrow{\mathrm{c}}=\alpha \overrightarrow{\mathrm{a}}+\beta \overrightarrow{\mathrm{b}} \ldots .(1) \\ & \overrightarrow{\mathrm{a}} . \overrightarrow{\mathrm{c}}=\alpha \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}+\beta \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}} \\ & 0=\alpha+\beta \cos 15^{\circ} \ldots .(2) \\ & (1) \Rightarrow \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\alpha \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}+\beta \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{~b}} \\ & \Rightarrow \cos 75^{\circ}=\alpha \cos 15^{\circ}+\beta \ldots .(3) \\ & (2) \&(3) \Rightarrow \cos 75^{\circ}=-\beta \cos ^2 15^{\circ}+\beta \end{aligned}$

$\begin{aligned} & \beta=\frac{\cos 75^{\circ}}{\sin ^2 15^{\circ}}=\frac{1}{\sin 15^{\circ}}=\frac{2 \sqrt{2}}{\sqrt{3}-1} \\ & (2) \Rightarrow \alpha=\frac{-\cos 15^{\circ}}{\sin 15^{\circ}}=\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)} \\ & \therefore \overrightarrow{\mathrm{c}}=\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)} \overrightarrow{\mathrm{a}}+\left(\frac{2 \sqrt{2}}{\sqrt{3}-1}\right) \overrightarrow{\mathrm{b}} \end{aligned}$

Now

$\begin{aligned} & \alpha+\sqrt{2}(\sqrt{3}-1) \beta=\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)}+\frac{\sqrt{2}(\sqrt{3}-1) \cdot 2 \sqrt{2}}{\sqrt{3}-1} \\ & =\frac{-(\sqrt{3}+1)^2}{2}+4 \\ & =\frac{-3-1-2 \sqrt{3}+8}{2} \\ & =2-\sqrt{3} \end{aligned}$

We are given two unit vectors $\vec{a}$ and $\vec{b}$ with an angle of $\frac{\pi}{3}$ between them. This means:

$\vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = 1$ (since they are unit vectors)

$\vec{a} \cdot \vec{b} = \cos\frac{\pi}{3} = \frac{1}{2}$

We need to find the number of values of $\lambda$ in the interval $[-1, 3]$ for which the vectors $\lambda \vec{a}+2 \vec{b}$ and $3 \vec{a}-\lambda \vec{b}$ are perpendicular, meaning their dot product is zero.

Here’s how to solve the problem step by step:

Set up the perpendicular condition by equating the dot product to zero:

$(\lambda \vec{a}+2 \vec{b})\cdot(3\vec{a}-\lambda \vec{b}) = 0.$

Expand the dot product using the distributive property:

$\begin{align*} (\lambda \vec{a}+2 \vec{b})\cdot(3\vec{a}-\lambda \vec{b}) &= \lambda \cdot 3 (\vec{a}\cdot\vec{a}) - \lambda^2 (\vec{a}\cdot\vec{b}) + 2\cdot 3 (\vec{b}\cdot\vec{a}) - 2\lambda (\vec{b}\cdot\vec{b})\\[1mm] &= 3\lambda - \lambda^2\left(\frac{1}{2}\right) + 6\left(\frac{1}{2}\right) - 2\lambda\\[1mm] &= 3\lambda - \frac{\lambda^2}{2} + 3 - 2\lambda. \end{align*}$

Simplify the expression:

$3\lambda - 2\lambda = \lambda,$

so the equation becomes:

$\lambda - \frac{\lambda^2}{2} + 3 = 0.$

Multiply the entire equation by 2 to eliminate the fraction:

$2\lambda - \lambda^2 + 6 = 0,$

which can be rearranged as:

$-\lambda^2 + 2\lambda + 6 = 0.$

Multiply both sides by -1 to get a standard quadratic form:

$\lambda^2 - 2\lambda - 6 = 0.$

Solve the quadratic equation using the quadratic formula:

$\lambda = \frac{2 \pm \sqrt{(-2)^2 - 4\cdot1\cdot(-6)}}{2} = \frac{2 \pm \sqrt{4+24}}{2} = \frac{2 \pm \sqrt{28}}{2}.$

Notice that:

$\sqrt{28} = 2\sqrt{7},$

thus:

$\lambda = \frac{2 \pm 2\sqrt{7}}{2} = 1 \pm \sqrt{7}.$

Evaluate the solutions numerically:

$\lambda = 1 + \sqrt{7} \approx 1 + 2.6458 \approx 3.6458,$

$\lambda = 1 - \sqrt{7} \approx 1 - 2.6458 \approx -1.6458.$

Check if the solutions are in the interval $[-1, 3]$:

$1 + \sqrt{7} \approx 3.6458$ is greater than 3.

$1 - \sqrt{7} \approx -1.6458$ is less than -1.

Since neither value lies within the interval $[-1, 3]$, there are no valid values of $\lambda$ in that interval.

Therefore, the number of values of $\lambda$ in the interval $[-1, 3]$ is:

0

$\begin{aligned} & \vec{u} \times \vec{v}=\vec{w} \text { and } \vec{v} \times \vec{w}=\vec{u} \\ & \Rightarrow \quad \vec{u}, \vec{v} \text { and } \vec{w} \text { are mutually perpendicular. } \\ & (\vec{v} \times \vec{w}) \times \vec{v}=\vec{w} \\ & \Rightarrow \quad \vec{w}(\vec{v} \cdot \vec{v})-\vec{v}(\vec{v} \cdot \vec{w})=\vec{w} \\ & \Rightarrow \quad \vec{w}\left(|\vec{v}|^2-1\right)-\vec{v}(\vec{v} \cdot \vec{w})=\overrightarrow{0}\end{aligned}$

$\begin{aligned} & \Rightarrow \quad|\vec{v}|^2=1 \text { and } \vec{v} \cdot \vec{w}=0 \\ & |\vec{u}||\vec{v}|=|\omega| \Rightarrow|\vec{u}|=\sqrt{6} \\ & \vec{u} \cdot \vec{w}=0 \Rightarrow \alpha+\beta-2 \gamma=0 \\ & \text { and }-t \alpha+\beta+\gamma=0 .......(i) \\ & \alpha-t \beta+\gamma=0 ........(ii) \\ & \alpha+\beta-t \gamma=0 ........(iii) \\ & \text { (ii) }- \text { (i) } \Rightarrow \quad \alpha(1+t)=(t+1) \beta \\ & \text { (iii) }- \text { (ii) } \Rightarrow \quad \beta(1+t)=(1+t) \gamma \\ & \Rightarrow \quad \alpha(1+t)=\beta(1+t)=\gamma(1+t)\end{aligned}$

either $t=-1$ or $\alpha=\beta=\gamma$

$ \begin{aligned} & \Rightarrow \quad \sqrt{\alpha^2+\alpha^2+\alpha^2}=\sqrt{6} \\ & \Rightarrow \quad \alpha=\sqrt{2},-t \alpha=-\alpha-\alpha \\ & \Rightarrow \quad t=2 \end{aligned} $

Since $\alpha=\sqrt{3}$

$ \begin{aligned} \Rightarrow & t=-1 \\ \Rightarrow & \alpha+\beta+\gamma=0 \\ & \alpha+\beta-2 \gamma=0 \\ \Rightarrow & \gamma=0 \end{aligned} $

$(\mathrm{Q}) \rightarrow 1$

$ \begin{aligned} & \alpha+\beta=0 \Rightarrow(\beta+\gamma)=-\alpha \Rightarrow(\beta+\gamma)^2=\alpha^2=3,(\mathrm{R}) \rightarrow 4 \\ & \Rightarrow|\vec{v}|^2=1 \rightarrow(\mathrm{P}) \rightarrow 2 \end{aligned} $

If $\alpha=\sqrt{3} \Rightarrow \gamma^2=0 \rightarrow(\mathrm{Q}) \rightarrow 1$

If $\alpha=\sqrt{3} \Rightarrow(\beta+\gamma)^2=(\sqrt{3})^2=3,(\mathrm{R}) \rightarrow 4$

If $\alpha=\sqrt{2}, t+3=(2)+3=5$, (S) $\rightarrow 5$

$ \begin{aligned} & \text { Let } \mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & \qquad \begin{aligned} \mathbf{b} & =2 \sqrt{3 \mathbf{i}}-2 \sqrt{3 \mathbf{j}}+\sqrt{3 \mathbf{k}} \\ \mathbf{c} & =-(2+2 \sqrt{3}) \hat{\mathbf{i}}+(2 \sqrt{3}-1) \hat{\mathbf{j}}+(2-\sqrt{3}) \hat{\mathbf{k}} \end{aligned} \\ & |\mathbf{a}|=3,|\mathbf{b}|=3 \sqrt{3},|\mathbf{c}|=6 \\ & \text { Perimeter }=9+3 \sqrt{3} \end{aligned} $

$ \begin{aligned} \cos A & =\frac{27+36-9}{36 \sqrt{3}} \\ \cos A & =\frac{54}{36 \sqrt{3}} \\ \cos A & =\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2} \\ A & =30^{\circ} \end{aligned} $

Let the normal vector of plane $\pi_1$ be

$ (\hat{i}+\hat{j}) \times(\hat{i}+2 \hat{j})=(2 \hat{k}-\hat{k})=\hat{k} $

Let the normal vector of plane $\pi_2$ be

$ (2 \hat{\mathbf{i}}-\hat{\mathbf{j}}) \times(3 \hat{\mathbf{i}}+2 \hat{\mathbf{k}})=-4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}-2 \hat{\mathbf{i}} $

Let the direction ratio of line along $a$ be $a, b$ and $c$

$ \begin{array}{ll} \Rightarrow \quad & a \times 0+b \times 0+c \times 1=0 \\ & -4 b+3 c-2 a=0 \\ & a \times 0+b \times 0+c \times 1=0 \\ & -2 \times a-4 \times b+c \times 3=0 \\ \Rightarrow \quad & \frac{a}{4}=\frac{-b}{2}=\frac{c}{0} \\ \Rightarrow \quad & \mathrm{a}=\frac{4 \hat{\mathrm{i}}}{2 \sqrt{5}}-\frac{2 \hat{\mathrm{j}}}{2 \sqrt{5}}+0 \hat{\mathrm{k}} \\ & \mathrm{~b}=\frac{\hat{\mathrm{i}}}{3}-\frac{2 \hat{\mathrm{j}}}{3}+\frac{2 \hat{\mathrm{k}}}{3} \\ & \mathrm{a} \cdot \mathrm{~b}=\cos \theta \\ & \cos \theta=\frac{4}{6 \sqrt{5}}+\frac{4}{6 \sqrt{5}}=\frac{4}{3 \sqrt{5}} \\ \Rightarrow \quad & \theta=\cos ^{-1}\left(\frac{4}{3 \sqrt{5}}\right) \end{array} $

$ \begin{aligned} &\begin{aligned} & \mathrm{AB}+\mathrm{BC}+\mathrm{CA}=0 \\ & \mathrm{BC}=\mathrm{AC}-\mathrm{AB} \end{aligned}\\ &\text { Let angle between } \mathbf{A B} \text { and } \mathbf{B C} \text { be }(\pi-\infty)\\ &\begin{aligned} & \mathbf{A B} \cdot \mathbf{B C}=|\mathbf{A B} \| \mathbf{B C}| \cos (\pi-\infty) \\ & \mathrm{AB} \cdot(\mathrm{AC}-\mathrm{AB})=-|\mathrm{AB} \| \mathrm{BC}| \cos \alpha \\ & \mathrm{AB} \cdot \mathrm{AC}-(\mathrm{AB})^2=-|\mathrm{AB} \| \mathrm{BC}| \cos \alpha \end{aligned} \end{aligned} $

$ \begin{aligned} & \mathbf{A B} \cdot \mathbf{A C}=-|\mathbf{A B} \| \mathbf{A C}| \times \frac{1}{2} \\ & \begin{aligned} |\mathbf{A B} \| \mathbf{A C}| & \times \frac{1}{2}-|\mathbf{A B}|^2 \\ & =-|\mathbf{A B} \| \mathbf{B C}| \cos \alpha \end{aligned} \end{aligned} $

$ \begin{aligned} & \frac{|\mathbf{A C}|}{|\mathbf{A B}|} \times \frac{1}{2}-1=\frac{-|\mathbf{B C}|}{|\mathbf{A B}|} \cos \alpha \\ & \Rightarrow \frac{-1}{10}+1=\frac{|\mathbf{B C}|}{|\mathbf{A B}|} \cos \alpha \\ & \Rightarrow \frac{9}{10}=\frac{|\mathbf{B C}|}{|\mathbf{A B}|} \cos \alpha \end{aligned} $

$ \Rightarrow|\mathbf{B C}|=\sqrt{|\mathbf{A C}|^2+|\mathbf{A B}|^2-} \begin{aligned} & 2|\mathbf{A C}||\mathbf{A B}| \times \frac{1}{2} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{|\mathbf{B C}|}{|\mathbf{A B}|}=\sqrt{\frac{|\mathbf{A C}|^2}{|\mathbf{A B}|^2}+1-\frac{|\mathbf{A C}|}{|\mathbf{A B}|}} \\ & \Rightarrow \frac{|\mathbf{B C}|}{|\mathbf{A B}|}=\sqrt{\frac{1}{25}+1-\frac{1}{5}} \\ & \Rightarrow \frac{|\mathbf{B C}|}{|\mathbf{A B}|}=\sqrt{\frac{1+25-5}{25}}=\frac{\sqrt{3} \cdot \sqrt{7}}{5} \\ & \Rightarrow \frac{9}{10}=\frac{\sqrt{3} \cdot \sqrt{7}}{5} \cos \alpha \Rightarrow \alpha=\cos ^{-1}\left(\frac{3 \sqrt{3}}{2 \sqrt{7}}\right) \end{aligned} $

$ \text { } \begin{aligned} & \mathbf{a} \times((\mathbf{r}-\mathbf{b}) \times \mathbf{a})+\mathbf{b} \times((\mathbf{r}-\mathbf{c}) \times \mathbf{b}) +\mathbf{c} \times((\mathbf{r}-\mathbf{a}) \times \mathbf{c})=\mathbf{0} \\ & k^2(3 \mathbf{r}-\mathbf{a}-\mathbf{b}-\mathbf{c})-(\mathbf{a} \cdot \mathbf{r}) \mathbf{a}-(\mathbf{b} \cdot \mathbf{r}) \mathbf{b} -(\mathbf{c} \cdot \mathbf{r}) \mathbf{c}=\mathbf{0} \end{aligned} $

$ \begin{array}{ll} \text { Let } \mathbf{r}= & x \mathbf{a}+y \mathbf{b}+z \mathbf{c} \\ & \mathbf{r} \cdot \mathbf{a}=k^2 x \\ & \mathbf{r} \cdot \mathbf{b}=k^2 y \\ & \mathbf{r} \cdot \mathbf{c}=k^2 z \\ \Rightarrow \quad & \mathbf{r}=\frac{(\mathbf{r} \cdot \mathbf{a}) \mathbf{a}+(\mathbf{r} \cdot \mathbf{b}) \mathbf{b}+(\mathbf{r} \cdot \mathbf{c}) \mathbf{c}}{k^2} \\ \Rightarrow \quad & 3 k^2 \mathbf{r}-k^2 \mathbf{r}=k^2(\mathbf{a}+\mathbf{b}+\mathbf{c}) \\ \Rightarrow \quad & \mathbf{r}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{2} \end{array} $

The shortest distance between two lines $\mathbf{r}=\mathbf{p}+t \mathbf{q}$ and $\mathbf{r}=\mathbf{c}+s \mathbf{d}$ is equal to $\frac{|(p-c) \cdot(q \times d)|}{|q \times d|}$, which is the projection of the vector ( $\mathbf{p}-\mathbf{c}$ ) on ( $\mathbf{q} \times \mathbf{d}$ ) $\mathbf{r}=\mathbf{a}+t \mathbf{b}$ and $\mathbf{r}=\mathbf{b}+s(\mathbf{a})$

Shortest distance between two lines

$ =\frac{|(\mathbf{a}-\mathbf{b}) \cdot(\mathbf{b} \times \mathbf{a})|}{|\mathbf{b} \times \mathbf{a}|}=\frac{|[\mathbf{a} \mathbf{b} \mathbf{a}]-[\mathbf{b} \mathbf{b} \mathbf{a}]|}{|\mathbf{b} \times \mathbf{a}|}=0 $

⇒ lines intersect

Thus, both A and R are true and R is the correct explanation of A .

$ \text { Let position vector of point } A \text { be } \mathbf{a} \text {. } $

Let position yector of point $B$ be $\mathbf{b}$.

$ \begin{aligned} \triangle B C D= & -\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\ \frac{\mathbf{b}+\mathbf{c}+\mathbf{d}}{3}= & -\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\ \mathbf{d}= & -3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-9 \hat{\mathbf{k}}-\mathbf{b}-\mathbf{c} \\ \mathbf{d}= & -3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-9 \hat{\mathbf{k}} \\ & -(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}+3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ = & -3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-9 \hat{\mathbf{k}}-\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ = & -4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-11 \hat{\mathbf{k}} \end{aligned} $

Let position vector of centroid $G$ of tetrahedron $A B C D$ be $\mathbf{g}$.

$ \begin{aligned} \mathbf{g} & =\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}}{4} \\ \mathbf{G D} & =\mathbf{d}-\left(\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}}{4}\right) \\ & =\frac{3 \mathbf{d}-|\mathbf{a}+\mathbf{b}+\mathbf{c}|}{4} \\ & =\frac{-12 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}-33 \hat{\mathbf{k}}-(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}})}{4} \\ & =\frac{-14 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}-38 \hat{\mathbf{k}}}{4} \\ & =\frac{-7 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-19 \hat{\mathbf{k}}}{2} \\ |\mathbf{G D}| & =\frac{1}{2} \times \sqrt{7^2+4^2+19^2} \\ & =\frac{\sqrt{2} \times \sqrt{213}}{\sqrt{2} \times \sqrt{2}}=\sqrt{\frac{213}{2}} \end{aligned} $

$ \begin{aligned} &\text { }|\mathbf{a}|=3|\mathbf{b}|=7\\ &\begin{aligned} |\mathbf{c}|= & 13 \\ \mid \mathbf{a}+\mathbf{b}+ & \left.\mathbf{c}\right|^2=|a|^2+|b|^2+|c|^2 \\ & +2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \\ = & 9+49+169+2((6-6-4) \\ & +(-24+9-24)+(-4-6+24) \\ = & 227+2(-4-39+14) \\ = & 227-58 \\ = & 169 \\ \Rightarrow \quad & |\mathbf{a}+\mathbf{b}+\mathbf{c}|=13 \\ \Rightarrow \quad & \sqrt{|\mathbf{a}|+|\mathbf{b}|+|\mathbf{c}|+|\mathbf{a}+\mathbf{b}+\mathbf{c}|} \\ = & \sqrt{10+13+13} \\ = & 6 \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} &|\mathbf{a}+2 \mathbf{b}|=|2 \mathbf{a}-\mathbf{b}| \\ &|\mathbf{a}+2 \mathbf{b}|^2=|2 \mathbf{a}-\mathbf{b}|^2 \\ &|a|^2+4|b|^2+4 \mathbf{a} \cdot \mathbf{b}=4|a|^2+|b|^2-4 a \cdot b \\ & \text { As }|\mathbf{a}|=|\mathbf{b}| \\ & \Rightarrow \mathbf{a} \cdot \mathbf{b}=0 \\ & \Rightarrow \mathbf{b} \cdot \mathbf{c}=0 \\ & \text { Angle }=90^{\circ} \end{aligned} $

$ \begin{aligned} &\text { Let angle between } \mathbf{a} \text { and } \mathbf{b} \text { be } \theta \text {. }\\ &\begin{aligned} |\mathbf{a}||\mathbf{b}| \cos \theta & =-1 \\ \cos \theta & =\frac{-1}{6} \\ |\mathbf{a} \times \mathbf{b}| & =|\mathbf{a}||\mathbf{b}| \sin \theta \\ & =\sqrt{6} \times \sqrt{6} \times \sqrt{1-\frac{1}{36}} \\ & =6 \times \frac{\sqrt{35}}{6} \\ & =\sqrt{35} \end{aligned} \end{aligned} $

$ \begin{aligned} |\mathbf{a} \times \mathbf{b}| \sin (\mathbf{a}, \mathbf{b})=\frac{\sqrt{35} \times \sqrt{35}}{6} & =\frac{35}{6} \\ \text { Also, }\left(|\mathbf{a}|^2-1\right)\left(1+\frac{1}{|\mathbf{b}|^2}\right) & =(5)\left(1+\frac{1}{6}\right) \\ & =\frac{35}{6} \end{aligned} $

$ \text { Let } \mathbf{A B}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} $

$ \begin{aligned} & \mathbf{A C}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\ & \mathbf{A D}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+p \hat{\mathbf{k}} \\ & \text { Volume }=\frac{1}{6}[\mathbf{A B} \mathbf{A C} \mathbf{A D}] \\ & \pm 12=\left|\begin{array}{rrr} 1 & 2 & -3 \\ 2 & 1 & -3 \\ 3 & -1 & p \end{array}\right| \\ & \pm 12=1(p-3)-2(2 p+9)-3(-2-3) \\ & \pm 12=p-3-4 p-18+15 \\ & \pm 12=-3 p-6 \\ & \Rightarrow \quad p=-6 \text { or } p=2 \\ & \Rightarrow p=2 \text { and }-6 \text { satisfies equation } \\ & x^2+4 x-12=0 \end{aligned} $

$ \begin{aligned} &\text { In } \triangle A B C \text {, }\\ &\begin{aligned} \mathbf{B C} & =\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \Rightarrow B C=\sqrt{9}=3 \\ \mathbf{C A} & =6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ \Rightarrow \quad C A & =\sqrt{36+9+4}=7 \\ \Rightarrow \quad \mathbf{A C} & =-6 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ \Rightarrow \quad \mathbf{A B} & =\mathbf{A C}-\mathbf{B C}=-7 \hat{\mathbf{i}}-\hat{\mathbf{j}} \\ \quad A B & =\sqrt{49+1}=\sqrt{50} \\ \text { Perimeter } & =3+7+\sqrt{50}=10+\sqrt{50} \\ & =10+5 \sqrt{2}=5(2+\sqrt{2}) \end{aligned} \end{aligned} $

Let position of $A, B, C$ and $D$ be $\mathbf{a}, \mathbf{b}$, c and d respectively.

$ \mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} $

In $\triangle B C D$ centroid is $\frac{\mathbf{b}+\mathbf{c}+\mathbf{d}}{3}$

$ \begin{aligned} \therefore \quad & \frac{\mathbf{b}+\mathbf{c}+\mathbf{d}}{3}=\frac{2}{3}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ \mathbf{b}+\mathbf{c}+\mathbf{d} & =2(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \end{aligned} $

∴ Centroid of tetrahedron $A B C D$ is

$ \begin{aligned} & =\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}}{4}=\frac{3(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})}{4} \\ \therefore \quad & \alpha=\frac{3}{4}, \beta=\frac{3}{4}, \gamma=\frac{3}{4} \\ \therefore \quad & 2 \alpha+\beta+\gamma=\frac{6+3+3}{4}=\frac{12}{4}=3 \end{aligned} $

$ \begin{aligned} & \text { } \mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \text { and } \\ & \qquad \begin{aligned} & \mathbf{b}=9 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-18 \hat{\mathbf{k}}=3(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}) \\ & \therefore \frac{\mathbf{b} \cdot \mathbf{a}}{\text { Projection of } \mathbf{b} \text { on } \mathbf{a}}=\frac{\frac{|\mathbf{a}|}{\mathbf{a} \cdot \mathbf{b}}}{|\mathbf{b}|}=\frac{|b|}{|a|} \\ &=\frac{3 \sqrt{9+4+36}}{\sqrt{1+4+4}}=7 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Let } \mathbf{r}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}\\ &\begin{aligned} & \therefore \quad \mathbf{r} \cdot \mathbf{a}=0 \\ & x+2 y+3 z=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \Rightarrow x=-2 y-3 z \\ & \text { And } \mathbf{r} \cdot \mathbf{b}=-2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \qquad \begin{aligned} & 2 x-3 y+z=-2 \\ & -4 y-6 z-3 y+z=-2 \\ & -5 z=-2+7 y \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & z=\frac{1}{5}(2-7 y) \\ & x=-2 y-\frac{3}{5}(2-7 y)=\frac{11 y-6}{5} \end{aligned} $

$ \begin{gathered}and\,\,\,\,\, \mathbf{r} \cdot \mathbf{C}=6 \\ 3 x+y-2 z=6 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{gathered} $

From by Eq. (iii), we get

$ \begin{array}{ll} \therefore & \frac{3}{5}(11 y-6)+y-\frac{2}{5}(2-7 y)=6 \\ \Rightarrow & 33 y-18+5 y-4+14 y=30 \\ \Rightarrow & 52 y=52 \\ & y=1 \\ & x=1 \\ \therefore & r \cdot(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & =3 x+y+z=3 \end{array} $