Vector Algebra

The position vectors are given as

$ \begin{aligned} & a=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, b=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & c=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}, d=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} $

$\because P$ divides $A B$ internally at the ratio $2: 1$

$ \begin{aligned} \therefore P & =\frac{2 b+a}{2+1}=\frac{2(\hat{i}+\hat{j}-2 \hat{k})+(\hat{i}-2 \hat{j}+\hat{k})}{3} \\ & =\hat{i}-\hat{k} \end{aligned} $

and $Q$ divides $C D$ externally in the $1: 2$

$ \begin{aligned} \therefore q & =\frac{\mathbf{d}-2 \mathbf{c}}{1-2}=\frac{(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})-2(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})}{-1} \\ & =3(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}) \end{aligned} $

Let $\mathbf{r}=5 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ divides $P Q$ in ratio $m: n$

$ =5 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} $

then, $\frac{n+3 m}{m+n}=5, \frac{-3 m}{m+n}=-6$

$ \frac{-n-3 m}{m+n}=-5 $

By solving these equations

We get, $m: n=-2: 1$

Hence, the ratio in which $R$ divides $P Q$ is $-2: 1$

$\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\mathbf{b}=2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$

$ \mathbf{a}=\mathbf{b} \times \mathbf{a} \text { and } \mathbf{r} \times \mathbf{b}=\mathbf{a} \times \mathbf{b} $

So, $\mathbf{r}=\mathbf{a}+\mathbf{b}$ satisfies both equations Thus $r=\mathbf{a}+\mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$

$ =\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}} $

then, $|r|=\sqrt{1^2+3^2+(-1)^2}=\sqrt{11}$

Therefore, the unit vector in the direction of

$ r=\frac{T}{|T|}=\frac{1}{\sqrt{11}}(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) $

$\because \mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\frac{\sqrt{3}}{2} \mathbf{b}+\frac{\mathbf{c}}{2}$

and $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}$

so, $(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}=\frac{\sqrt{3}}{2} \mathbf{b}+\frac{\mathbf{c}}{2}$

therefore, $\mathbf{a} \cdot \mathbf{c}=\frac{\sqrt{3}}{2}$

Which gives $\alpha=\pi / 6$ because $\mathbf{a}$ and $\mathbf{c}$ are unit vector

and $\mathbf{a} \cdot \mathbf{b}=-\frac{1}{2}$

which gives $\beta=\frac{2 \pi}{3}$

Hence, $\alpha+\beta=\frac{\pi}{6}+\frac{2 \pi}{3}=\frac{5 \pi}{6}$

In $\triangle A B C$,

$A(\mathbf{a}), B(\mathbf{b})$ and $C(\mathbf{c})$ are the vertices and circumcentre is $P\left(\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{4}\right)$.

Centroid of $\triangle A B C, \mathbf{G}=\left(\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}\right)$

$ \begin{aligned} &\text { ∴ Position vector of orthocentre }\\ &\begin{aligned} \frac{2\left(\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{4}\right)+\mathbf{H}}{3} & =\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3} \\ \mathbf{H} & =\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{2} \end{aligned} \end{aligned} $

We have,

$ \begin{aligned} & \mathbf{O A}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})=(1,2,2) \\ & \mathbf{O B}=(2 \hat{\mathbf{i}}-\hat{\mathbf{j}})=(2,-1,0) \\ & \mathbf{O C}=(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=(1,1,3) \end{aligned} $

and $\mathbf{O D}=(4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})=(0,4,5)$

$ \begin{aligned} A B & =\sqrt{1+9+4}=\sqrt{14} \\ B C & =\sqrt{1+4+9}=\sqrt{14} \\ C D & =\sqrt{1+9+4}=\sqrt{14} \\ A D & =\sqrt{1+4+9}=\sqrt{14} \\ A C & =\sqrt{0+1+1}=\sqrt{2} \\ B D & =\sqrt{1+4+9}=\sqrt{14} \\ \because \quad A B & =B C=C D=D A \text { and diagonal } \\ A C & \neq B D \end{aligned} $

$\therefore A B C D$ is rhombus

$ \begin{aligned} &\text { ∴ Angle betweem } \mathbf{a} \text { and } \mathbf{b} \text { is obtuse }\\ &\begin{aligned} & \therefore \cos \theta<0 \\ & \mathbf{a} \cdot \mathbf{b}<0 \\ & \begin{aligned} (c x \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+ & 3 \hat{\mathbf{k}}) \cdot(x \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2(c x \hat{\mathbf{k}})<0 \\ & =c x^2-12+6 c x<0 \\ & =c x^2+6 c x-12<0, \forall x \in R \end{aligned} \end{aligned} \end{aligned} $

$ \begin{gathered} \therefore c<0 \text { and } D<0 \\ 36 c^2-4 c(-12)<0 \\ 3 c^2+4 c<0 \\ c(3 c+4)<0 \end{gathered} $

$ \begin{aligned} & \therefore c \in\left(\frac{-4}{3}, 0\right), c^{\prime} \text { can be zero } \\ & \therefore c \in\left(-\frac{4}{3}, 0\right] \end{aligned} $

$ \text { } \begin{aligned} & \mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \mathbf{b}=3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \mathbf{c}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \because \quad \mathbf{r} \times \mathbf{a}=\mathbf{r} \times \mathbf{b} \\ & \mathbf{r} \times(\mathbf{a}-\mathbf{b})=\mathbf{0} \\ & \therefore \quad \mathbf{r} \|(\mathbf{a}-\mathbf{b}) \\ & \mathbf{r}=\lambda(\mathbf{a}-\mathbf{b}) \\ & \because \quad \mathbf{r} \cdot \mathbf{c}=18 \\ & \Rightarrow \lambda(\mathbf{a} \cdot \mathbf{c}-\mathbf{b} \cdot \mathbf{c})=18 \\ & \Rightarrow \lambda(2-2+9-(3-6+3)]=18 \\ & \Rightarrow \quad \lambda \lambda=18 \\ & \Rightarrow \quad \lambda=2 \\ & \mathbf{r}=2(-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} $

$ \begin{aligned} \cos \theta & =\left\lvert\, \frac{(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \cdot 2(-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{k})}{\sqrt{16+9+1} 2 \sqrt{1+4+4}}\right| \\ & =\frac{12}{\sqrt{26} \cdot 3}=\frac{4}{\sqrt{26}} \end{aligned} $

$ \text { ∴ } \text { Projector }=\sqrt{16+9+1} \cdot \frac{4}{\sqrt{26}}=4 $

∵ $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are non-coplaner vectors,

$ \Rightarrow[\mathbf{u} \vee \mathbf{v}] \neq 0 $

Now, according to the question,

$\therefore \quad\left[\begin{array}{lll}3 \mathbf{u} & p \mathbf{v} & p \mathbf{w}\end{array}\right]-\left[\begin{array}{lll}p \mathbf{v} & \mathbf{w} & q u\end{array}\right]$

$-\left[\begin{array}{lll}2 w & q v & q u\end{array}\right]=0$

$\Rightarrow \quad 3 p^2\left[\begin{array}{lll}\mathbf{u} & \mathbf{v} & \mathbf{w}\end{array}\right]-p q\left[\begin{array}{lll}\mathbf{v} & \mathbf{w} & \mathbf{u}\end{array}\right]$

$-2 q^2\left[\begin{array}{lll}\mathbf{w} & \mathbf{v} & \mathbf{u}\end{array}\right]=0$

$\Rightarrow \quad\left[\begin{array}{lll}\mathbf{u} & \mathbf{v} & \mathbf{w}\end{array}\right]\left(3 p^2-p q-2 q^2\right)=0$

$\Rightarrow \quad 3 p^2-p q-2 q^2=0$

$\Rightarrow \quad 3 p^2-3 p q+2 p q-2 q^2=0$

$\Rightarrow \quad(3 p+2 q)(p-q)=0$

$\therefore \quad p=q,-\frac{2 q}{3}$

Step 1: List the Vectors

The two vectors are:
$\mathbf{a} = x \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - \hat{\mathbf{k}}$
$\mathbf{b} = 6 \hat{\mathbf{i}} - y\hat{\mathbf{j}} + 2 \hat{\mathbf{k}}$

Step 2: Use the Given Relationship

We are given that:
$|\mathbf{a} \times \mathbf{b}|^2 + |\mathbf{a} \cdot \mathbf{b}|^2 = f(x) g(y)$

Step 3: Recognize the Identity

There is a formula: $|\mathbf{a} \times \mathbf{b}|^2 + |\mathbf{a} \cdot \mathbf{b}|^2 = |\mathbf{a}|^2 \cdot |\mathbf{b}|^2$

Step 4: Find the Lengths of the Vectors

For $\mathbf{a}$:
$|\mathbf{a}| = \sqrt{x^2 + 2^2 + (-1)^2} = \sqrt{x^2 + 4 + 1} = \sqrt{x^2 + 5}$
For $\mathbf{b}$:
$|\mathbf{b}| = \sqrt{6^2 + y^2 + 2^2} = \sqrt{36 + y^2 + 4} = \sqrt{y^2 + 40}$

Step 5: Substitute in the Formula

$ |\mathbf{a}|^2 \cdot |\mathbf{b}|^2 = (x^2 + 5)(y^2 + 40) $

So, $ f(x)g(y) = (x^2 + 5)(y^2 + 40) $

Step 6: Identify $f(x)$ and $g(y)$

So, $ f(x) = x^2 + 5 $

$ g(y) = y^2 + 40 $

Step 7: Use the Given Equation

$f(x) + g(y) - 46 = 0$

Step 8: Substitute for $f(x)$ and $g(y)$

$x^2 + 5 + y^2 + 40 - 46 = 0$
$x^2 + y^2 - 1 = 0$

Step 9: Final Form

So the equation is:
$x^2 + y^2 = 1$

We have, $\mathbf{r}=\mathbf{a} \times \mathbf{b}-\mathbf{c} \times \mathbf{b}-\mathbf{a} \times \mathbf{c}$

$ \begin{aligned} & \Rightarrow \mathbf{r}=\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a} \\ & \Rightarrow \mathbf{r} \cdot \mathbf{a}=\alpha \cdot \mathbf{l}, \mathbf{r} \cdot \mathbf{b}=\alpha, \mathbf{r} \cdot \mathbf{c}=\alpha \end{aligned} $

Equation of plane passing through $A(\mathbf{a}), B(\mathbf{b})$ and $C(\mathbf{c})$ is

$ \begin{aligned} & n=A B \times B C \\ & =(b-a) \times(c-b)=b \times c-a \times c+a \times b \\ & =a \times b+b \times c+c \times a=r \end{aligned} $

Equation of plane

$ \quad \begin{aligned} & \left(\mathbf{r}_1-\mathbf{a}\right) \cdot \mathbf{n}=0 \\\Rightarrow & \mathbf{r}_1 \cdot \mathbf{n}=\mathbf{a} \cdot \mathbf{n} \Rightarrow \mathbf{r}_1 \cdot \mathbf{r}=\mathbf{a} \cdot \mathbf{r} \\ & \mathbf{r}_1 \cdot \mathbf{r}=\alpha \\ & \mathbf{r}_1 \cdot \frac{\mathbf{r}}{|\mathbf{r}|}=\frac{|\alpha|}{|\mathbf{r}|} \Rightarrow \mathbf{r} \cdot \hat{\mathbf{n}}=p \end{aligned} $

$\therefore \frac{|\alpha|}{|\mathbf{r}|}$ is perpendicular distance from origin to plane.

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

$ \begin{array}{ll} \therefore & \mathbf{a} \times \hat{\mathbf{i}}=-b \hat{\mathbf{k}}+c \hat{\mathbf{j}} \Rightarrow \mathbf{a} \times \hat{\mathbf{j}}=a \hat{\mathbf{k}}-c \hat{\mathbf{i}} \\ \Rightarrow & \mathbf{a} \times \hat{\mathbf{k}}=-a \hat{\mathbf{j}}+b \hat{\mathbf{i}} \\ \therefore & (\mathbf{a} \times \hat{\mathbf{i}})^2=b^2+c^2 \\ & (\mathbf{a} \times \hat{\mathbf{j}})^2=a^2+c^2 \\ & (\mathbf{a} \times \hat{\mathbf{k}})^2=a^2+b^2 \\ \because & P=2\left(a^2+b^2+c^2\right) \end{array} $

Now, $Q=a^2+b^2+c^2$

$ \therefore \quad P=2 Q $

$ \begin{aligned} &\text { A Given, }\\ &\begin{aligned} & \mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \\ & \mathbf{c}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \mathbf{r} \cdot \mathbf{a}=0 \Rightarrow[\mathbf{r} \cdot \mathbf{a} \mathbf{b}]=0 \Rightarrow \mathbf{r} \cdot(\mathbf{a} \times \mathbf{b})=0 \\ & \therefore \quad \mathbf{r}=\lambda(\mathbf{a} \times(\mathbf{a} \times \mathbf{b}) \\ & =\lambda[(\mathbf{a} \cdot \mathbf{b}) \mathbf{a}-\mathbf{a} \cdot \mathbf{a}(\mathbf{b})] \\ & =\lambda(9 \mathbf{a}-6 \mathbf{b})=3 \lambda(3 \mathbf{a}-2 \mathbf{b}) \\ & \mathbf{r}=k(3 \mathbf{a}-2 \mathbf{b}) \\ & \mathbf{r} \cdot \mathbf{c}=k(3 \mathbf{a} \cdot \mathbf{c}-2 \mathbf{b} \cdot \mathbf{c}) \\ & \Rightarrow \quad 3=k[3(2-(-2)-2(2-2-3)] \\ & \Rightarrow \quad 3=k(-3+6) \Rightarrow k=7 \\ & \therefore \quad \mathbf{r}=\hat{\mathbf{i}}-\hat{\mathbf{j}} \Rightarrow|\mathbf{r}|=\sqrt{2} \end{aligned} \end{aligned} $

The position vector of $\mathbf{a}$ is

$ \mathbf{a}=-3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} $

Let, $M$ be the mid-point of the hypotenuse $B C$.

So, the position vector of $\mathbf{M}$ is

$ \mathbf{m}=6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}} $

Let the centroid be $G$ and its position vector be $\mathbf{g}$.

So, $\mathbf{g}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Since, the mid-point of the hypotenuse of a right triangle is equidistant from all three vertices.

So, $M$ is the circumcenter of a triangle and $\mathbf{m}=\frac{\mathbf{b}+\mathbf{c}}{2}$

$\Rightarrow \mathbf{b}+\mathbf{c}=2 \mathbf{m}$

Substituting this value in Eq. (i), we get

$ \begin{aligned} &\begin{aligned} g & =\frac{\mathbf{a}+2 \mathbf{m}}{3} \\ \Rightarrow g & =\frac{(-3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})+2(6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})}{3} \\ & =\frac{-3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}+12 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}}{3} \\ & =\frac{9 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}}{3} \\ & =3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \end{aligned}\\ &\text { So, the position vector of centroid is }\\ &3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \end{aligned} $

Given,

$ \begin{aligned} & \mathbf{A}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{B}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}} \text { and } \\ & \begin{aligned} \mathbf{C}= & 5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+5 \hat{\mathbf{k}} \\ \text { So, } \mathbf{B C} & =\mathbf{C}-\mathbf{B} \\ & =5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}-\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & =4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \end{aligned} \\ & \begin{aligned} \mathbf{B A}= & \mathbf{A}-\mathbf{B}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}}-\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ = & 2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \end{aligned} \end{aligned} $

Now, $\mathbf{B A} \times \mathbf{B C}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & -2 \\ 4 & 2 & 4\end{array}\right|$

$ \begin{aligned} & =(4+4) \hat{\mathbf{i}}-(8+8) \hat{\mathbf{j}}+(4-4) \hat{\mathbf{k}} \\ & =8 \hat{\mathbf{i}}-16 \hat{\mathbf{j}} \end{aligned} $

And $|\mathbf{B A} \times \mathbf{B C}|=\sqrt{8^2+(-16)^2}$

$ \begin{aligned} & =\sqrt{64+256} \\ & =\sqrt{320}=8 \sqrt{5} \end{aligned} $

$ \begin{aligned} & \text { So, area of triangle }=\frac{1}{2}|\mathbf{B A} \times \mathbf{B C}| \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{2} \times 8 \sqrt{5}=4 \sqrt{5} \\ & \text { } \begin{aligned} Now,\,\, |\mathbf{B C}| & =\sqrt{4^2+2^2+4^2} \\ & =\sqrt{16+4+16}=\sqrt{36}=6 \end{aligned} \end{aligned} $

Let, the magnitude of the altitude $h$ from $A$ to $B C$ is given by

$ \begin{aligned} & \text { Area }=\frac{1}{2} \times \text { base } \times \text { height } \\ & \Rightarrow \quad 4 \sqrt{5}=\frac{1}{2} \times|B C| \times h \\ & \Rightarrow \quad 4 \sqrt{5}=\frac{1}{2} \times 6 \times h \\ & \Rightarrow \quad h=\frac{4 \sqrt{5}}{3} \end{aligned} $

Let $\mathbf{a}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$,

$ \mathbf{b}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \text { and } \mathbf{c}=p \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} $

Now, $\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=\left|\begin{array}{ccc}2 & 4 & -3 \\ -1 & 2 & 3 \\ p & -2 & 1\end{array}\right|=0$

$ \begin{aligned} & \Rightarrow \quad 2(2+6)-4(-1-3 p)-3(2-2 p)=0 \\ & \Rightarrow \quad 16+4+12 p-6+6 p=0 \\ & \Rightarrow \quad 18 p+14=0 \\ & \Rightarrow \quad p=\frac{-14}{18}=\frac{-7}{9} \end{aligned} $

Now, $\mathbf{v}=9 p \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$

$ \begin{aligned} & =9\left(\frac{-7}{9}\right) \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\ & =-7 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \end{aligned} $

Now, unit vector

$ \begin{aligned} \hat{\mathbf{u}} & =\frac{\mathbf{v}}{|\mathbf{v}|}=\frac{-7 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}}{\sqrt{(-7)^2+(-4)^2+4^2}} \\ \Rightarrow \quad \hat{\mathbf{u}} & =\frac{-7 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}}{9} \\ & =\frac{-7}{9} \hat{\mathbf{i}}-\frac{4}{9} \hat{\mathbf{j}}+\frac{4}{9} \hat{\mathbf{k}} \end{aligned} $

Let $\mathbf{b}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}$ and $\mathbf{a}$ is perpendicular to $\mathbf{b}$.

$ \begin{array}{ll} \therefore & \mathbf{a} \cdot \mathbf{b}=0 \\ \Rightarrow & (4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \cdot(x \hat{\mathbf{i}}+y \hat{\mathbf{j}})=0 \end{array} $

$ \begin{array}{ll} \Rightarrow & 4 x+3 y=0 \\ \Rightarrow & \frac{x}{3}=\frac{y}{-4}=\lambda \\ \Rightarrow & x=3 \lambda, y=-4 \lambda \\ \therefore & \mathbf{b}=\lambda(3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}) \end{array} $

Let the required vector be $\mathbf{c}=c_1 \hat{\mathbf{i}}+c_2 \hat{\mathbf{j}}$ then the projection of $\mathbf{c}$ on $\mathbf{a}$ and $b\,\,{\text {are }} \frac{\mathbf{c} \cdot \mathbf{a}}{|\mathbf{a}|}$ and $\frac{\mathbf{c} \cdot \mathbf{b}}{|\mathbf{b}|}$ respectively.

$ \begin{aligned} & \therefore \quad \frac{\mathbf{c} \cdot \mathbf{a}}{|\mathbf{a}|}=1 \text { and } \frac{\mathbf{c} \cdot \mathbf{b}}{|\mathbf{b}|}=2 \\ & \Rightarrow \quad \frac{\left(c_1 \hat{\mathbf{i}}+c_2 \hat{\mathbf{j}}\right) \cdot(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}})}{\sqrt{4^2+3^2}}=1 \\ & \Rightarrow \quad 4 c_1+3 c_2=5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { and } \quad \frac{\mathbf{c} \cdot \mathbf{b}}{|\mathbf{b}|}=2 \\ & \Rightarrow \quad \frac{\left(c_1 \hat{\mathbf{i}}+c_2 \hat{\mathbf{j}}\right) \cdot(3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}})}{\sqrt{3^2+(-4)^2}}=2 \\ & \Rightarrow \quad 3 c_1-4 c_2=10 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Solving Eq. (i) and (ii), we get

$ c_1=2, c_2=-1 $

So, the required vector

$ =c_1 \hat{\mathbf{i}}+c_2 \hat{\mathbf{j}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}} $

$\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}, \mathbf{b}=-\hat{\mathbf{i}}+3 \mathbf{j}+3 \hat{\mathbf{k}}$

$ =3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} $

and $|\mathbf{a}-\mathbf{b}|=\sqrt{3^2+(-6)^2+2^2}=7$

Unit vector in (a - b)

$ =\frac{1}{7}(3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) $

thus, the vector of magnitude 28 units

$ \begin{aligned} & =28\left(\frac{1}{7}(3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})=4(3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\right) \\ & =12 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+8 \hat{\mathbf{k}} \end{aligned} $

Let, $\mathbf{a}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$

$ \begin{aligned}then, \mathbf{a} \times \mathbf{i} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ x & y & z \\ 1 & 0 & 0 \end{array}\right| \\ & =\hat{\mathbf{i}}(0-0)-\hat{\mathbf{j}}(0-z)+\hat{\mathbf{k}}(0-y) \\ & =z \hat{\mathbf{j}}-y \hat{\mathbf{k}} \end{aligned} $

So, $|\mathbf{a} \times \mathbf{i}|^2=z^2+y^2$

Similarly, $|\mathbf{a} \times \hat{\mathbf{j}}|^2=x^2+z^2$

and $|\mathbf{a} \times \hat{\mathbf{k}}|^2=x^2+y^2$

thus, $|\mathbf{a} \times \hat{\mathbf{i}}|^2+|\mathbf{a} \times \hat{\mathbf{j}}|^2+|\mathbf{a} \times \hat{\mathbf{k}}|^2$

$ =2\left(x^2+y^2+z^2\right) $

$\because \mathbf{a}$ is unit vector

So, $x^2+y^2+z^2=1$

Hence, $|\mathbf{a} \times \hat{\mathbf{i}}|^2+|\mathbf{a} \times \hat{\mathbf{j}}|^2+|\mathbf{a} \times \hat{\mathbf{k}}|^2$

$ =2 \times 1=2 $

$ \text {} \begin{aligned} (\mathbf{a} \times \mathbf{b}) & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -2 & -3 \\ -2 & 3 & 4 \end{array}\right| \\ = & \hat{\mathbf{i}}(-8+9)-\hat{\mathbf{j}}(4-6)+\hat{\mathbf{k}}(3-4) \\ = & \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ \text { and }(\mathbf{c} \times \mathbf{d}) & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 5 & -4 & 3 \\ 3 & 1 & 5 \end{array}\right| \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\hat{\mathbf{i}}(-20-3)-\hat{\mathbf{j}}(25-9)+\hat{\mathbf{k}}(5+12) \\ & =-23 \hat{\mathbf{i}}-16 \hat{\mathbf{j}}+17 \hat{\mathbf{k}} \end{aligned}\\ &\begin{aligned} & \text { Thus, }(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & - \\ -23 & -16 & 7 \end{array}\right| \\ & =\hat{\mathbf{i}}(34-16)-\hat{\mathbf{j}}(17-23)+\hat{\mathbf{k}}(-16+4) \\ & =18 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+30 \hat{\mathbf{k}} \end{aligned} \end{aligned} $

$ \begin{aligned} \mathbf{A B}=\mathbf{O B}-\mathbf{O A}= & (2-3) \hat{\mathbf{i}}+(0-1) \hat{\mathbf{j}} +(1-1) \hat{\mathbf{k}}=-\hat{\mathrm{i}}-\hat{\mathbf{j}} \end{aligned} $

So, the direction vector of $\operatorname{lin} \epsilon A B$ is

$ \begin{aligned} & \begin{aligned} \mathbf{d}= & -\hat{\mathbf{i}}-\hat{\mathbf{j}} \\ \text { let } \mathbf{P}= & \mathbf{A}+\lambda \mathbf{d} \\ = & (3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+\lambda(-\hat{\mathbf{i}}-\hat{\mathbf{j}}) \\ = & (3-\lambda) \hat{\mathbf{i}}+(1-\lambda) \hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} \\ & \text { and } \mathbf{P C}=\mathbf{C}-\mathbf{P}=(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}) \\ & \qquad \begin{aligned} - & {[(3-\lambda) \hat{\mathbf{i}}+(1-\lambda) \hat{\mathbf{j}}+\hat{\mathbf{k}}) } \\ & =(\lambda-2 \hat{\mathbf{i}}+(4+\lambda) \hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} \end{aligned} $

As, given in the question

$ \begin{aligned} & \because \mathrm{PC} \perp \mathrm{AB} \\ & \Rightarrow \mathrm{PC} \cdot \mathrm{AB}=0 \\ & \Rightarrow(\lambda-2)(-1)+(4+\lambda)(-1)+(-1)(0)=0 \\ & \Rightarrow-\lambda+2-4-\lambda=0 \\ & \Rightarrow \lambda=-1 \\ & \begin{aligned} \therefore \quad \mathrm{P} & =(3-\lambda) \hat{\mathrm{i}}+(1-\lambda) \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & =(3+1) \hat{\mathrm{i}}+(1+1) \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & =4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \end{aligned} \end{aligned} $

therefore, $a=4, b=2$ and $c=1$

Hence, $a+b+c=4+2+1=7$

$ \begin{aligned} \text { Let } \mathbf{a} & =2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+l \hat{\mathbf{k}} \\ \mathbf{b} & =-3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-4 l \hat{\mathbf{k}} \\ \mathbf{c} & =\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 l \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} &\text { For right angled } \Delta \text {; }\\ &\begin{aligned} & |\mathbf{b}|^2=|\mathbf{a}|^2+|\mathbf{c}|^2 \\ \Rightarrow \quad & \left(9+4+16 l^2\right)=\left(4+9+l^2\right) +\left(1+1+9 l^2\right) \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} \Rightarrow & l^2=\frac{1}{3} \Rightarrow l= \pm \frac{1}{\sqrt{3}} \text { but } l>0 \\ \therefore & l=\frac{1}{\sqrt{3}} \end{array}\\ &\text { So, its hypotenuse } \mathbf{b}=-3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\frac{4}{\sqrt{3}} \hat{\mathbf{k}}\\ &\begin{aligned} \therefore \quad|\mathbf{b}| & =\sqrt{9+4+\frac{16}{3}} \\ & =\sqrt{\frac{27+12+16}{3}}=\sqrt{\frac{55}{3}} \end{aligned} \end{aligned} $

Let $\mathbf{a}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$,

$ \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{c}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} $

So, a vector which is perpendicular to a and coplanar with $\mathbf{b} \times \mathbf{c}=\mathbf{a} \times(\mathbf{b} \times \mathbf{c})$

$ \begin{aligned} & =(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \\ & =0 \cdot \mathbf{b}-(-1) \mathbf{c}=\mathbf{c}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} \text { So, its unit vector } & =\frac{2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}}{\sqrt{2^2+2^2+(-1)^2}} \\ & =\frac{2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}}{3} \end{aligned} $

Since, given vectors are coplanar.

So, its determinant value $=0$.

$ \begin{aligned} & \Rightarrow\left|\begin{array}{ccc} 2 & -1 & 3 \\ 1 & 4 & 1 \\ 4 & p & 1 \end{array}\right|=0 \\ & \Rightarrow \quad 2(4-p)+1(1-4)+3(p-16)=0 \\ & \Rightarrow \quad p=43 \end{aligned} $

(Logically) given that $|\mathbf{a}|=3$,

$ |\mathbf{b}|=4,|\mathbf{a}+\mathbf{b}|=5 $

So, $|\mathbf{a}-\mathbf{b}|=5$

Direction vector of line $A B$ is

$ \begin{aligned} & \mathbf{A B}=\mathbf{O B}-\mathbf{O A}=(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}) -(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned} $

Similarly, direction vector of line $C D$ is

$ \begin{aligned} & \mathrm{CD}=\mathrm{OD}-\mathrm{OC}=(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})-(\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ & =2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}} \end{aligned} $

and $\mathbf{A C}=\mathbf{O C}-\mathbf{O A}=-\hat{\mathbf{i}}+0 \cdot \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$

$\therefore$ Shortest distance (S.D)

$ \begin{aligned} & =\left|\frac{\mathbf{A C} \cdot(\mathbf{A B} \times \mathbf{C D})}{|\mathbf{A B} \times \mathbf{C D}|}\right| \\ \mathbf{A B} \times \mathbf{C D} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -2 & 1 & 2 \\ 2 & -2 & 0 \end{array}\right| \\ & =4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned} $

and $\mathbf{A C} \cdot(\mathbf{A B} \times \mathbf{C D})=2$

Also, $|\mathbf{A B} \times \mathbf{C D}|=\sqrt{16+16+4}=4$

Hence, S.D. $=\left|\frac{2}{6}\right|=\frac{1}{3}$

Logically, the vector must be proportional to $(3,-2,6)$ with magnitude 63.

So, $P Q=9(3,-2,6)=(27,-18,54) \rightarrow$ Component form.

The lines makes an obtuse angle with the $X$-axis.

$\therefore$ Components of the vector $P Q$ are $-27,18,-54$

$ 4 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, 6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}} $

$\Rightarrow$ points are $P(4,1,3), Q(6,-2,-3)$ and $R(1,-1,-3)$

$ \begin{aligned} &\begin{aligned} \Rightarrow P Q & =\sqrt{(6-4)^2+(-3)^2+(-6)^2} \\ & =\sqrt{4+9+36}=\sqrt{49}=7 \\ \Rightarrow Q R & =\sqrt{(1-6)^2+(-1+2)^2+(-3+3)^2} \\ & =\sqrt{(-5)^2+(1)^2+0}=\sqrt{26} \\ \Rightarrow P R & =\sqrt{(1-4)^2+(-1-1)^2+(-3-3)^2} \\ & =\sqrt{9+4+36}=\sqrt{49}=7 \\ \Rightarrow P Q & =P R=7 \text { units } \end{aligned}\\ &\Rightarrow \triangle P Q R \text { is an isosceles triangle. } \end{aligned} $

$ \begin{aligned} &\text { Magnitude of the vector }-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\\ &\begin{aligned} & \Rightarrow|\mathbf{b}|=\sqrt{4+1+4}=\sqrt{9}=3 \\ & \Rightarrow \text { Unit vector } \hat{\mathbf{b}}=\frac{-2 \hat{\mathbf{i}}}{3}-\frac{1}{3} \hat{\mathbf{j}}+\frac{2}{3} \hat{\mathbf{k}} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Now, let } \mathbf{c}=\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\\ &\begin{aligned} & |\mathbf{c}|=\sqrt{54}=3 \sqrt{6} \\ & \hat{\mathbf{c}}=\frac{1}{3 \sqrt{6}} \hat{\mathbf{i}}-\frac{7}{3 \sqrt{6}} \hat{\mathbf{j}}+\frac{2}{3 \sqrt{6}} \hat{\mathbf{k}} \\ & \Rightarrow \mathbf{c}=\mathbf{a}+\mathbf{b} \Rightarrow \mathbf{a}=\mathbf{c}-\mathbf{b} \\ & =\frac{(1+2 \sqrt{6}) \hat{\mathbf{i}}}{3 \sqrt{6}}+\frac{(-7+\sqrt{6}) \hat{\mathbf{j}}}{3 \sqrt{6}}+\frac{(2-2 \sqrt{6}) \hat{\mathbf{k}}}{3 \sqrt{6}} \\ & \text { Compaire with } \mathbf{a}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ & \Rightarrow x=\frac{1+2 \sqrt{6}}{3 \sqrt{6}}=\frac{\sqrt{6}+12}{18} \cong \frac{7}{9} \\ & \Rightarrow x=\frac{7}{9} \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} & \mathbf{a}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+6 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \\ & \mathbf{c}=3 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & \mathbf{a} \times \mathbf{b}=\left|\begin{array}{rrr} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & -1 & 6 \\ 1 & -1 & 1 \end{array}\right| \\ &=\hat{\mathbf{i}}(-1+6)-\hat{\mathbf{j}}(2-6)+\hat{\mathbf{k}}(-2+1) \\ &=5 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & \mathbf{b} \times \mathbf{c}=\left|\begin{array}{rrr} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 1 \\ 0 & 3 & -1 \end{array}\right| \\ &=\hat{\mathbf{i}}(1-3)-\hat{\mathbf{j}}(-1-0)+\hat{\mathbf{k}}(3-0) \\ &=-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} \mathbf{c} \times \mathbf{a} & =\left|\begin{array}{rrr} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & 3 & -1 \\ 2 & -1 & 6 \end{array}\right| \\ & =\hat{\mathbf{i}}(18-1)-\hat{\mathbf{j}}(+2)+\hat{\mathbf{k}}(-6) \\ & =17 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \\ \mathbf{a} \times \mathbf{b} & +\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a} \\ & =5 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}}+(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})+17 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \\ & =20 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \end{aligned} $

$\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$

$ \mathbf{a} \cdot \mathbf{c}=|\mathbf{c}|,|\mathbf{c}-\mathbf{a}|=2 \sqrt{2} $

$\mathbf{a} \times \mathbf{b}$ and $\mathbf{c}$ is $30^{\circ}$.

$ \begin{aligned} & |(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=|\mathbf{a} \times \mathbf{b}||\mathbf{c}| \sin \theta \\ & \quad=|\mathbf{a} \times \mathbf{b}||\mathbf{c}| \times \frac{1}{2} \\ & \mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}} \\ & \mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{array}\right|=\hat{\mathbf{i}}(2)-\hat{\mathbf{j}}(2)+\hat{\mathbf{k}}(1) \\ & \quad=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & |\mathbf{a} \times \mathbf{b}|=\sqrt{4+4+1}=3 \\ & \quad \mathbf{a} \cdot \mathbf{c}=|\mathbf{c}| \\ & \Rightarrow|\mathbf{a} \| \mathbf{c}| \cos \alpha=|\mathbf{c}| \end{aligned} $

$ \begin{aligned} & \cos \alpha=\frac{1}{3} \\ & |\mathbf{c}-\mathbf{a}|=2 \sqrt{2} \\ & \Rightarrow|\mathbf{c}-\mathbf{a}|^2=8 \\ & \Rightarrow|\mathbf{c}|^2+|\mathbf{a}|^2-2|\mathbf{c}||\mathbf{a}| \cos \alpha=8 \\ & \Rightarrow|\mathbf{c}|^2+9-2|\mathbf{c}| 3 \times \frac{1}{3}=8 \\ & \Rightarrow|\mathbf{c}|^2-2|\mathbf{c}|+1=0 \\ & \Rightarrow(|\mathbf{c}|-1)^2=0 \Rightarrow|\mathbf{c}|=1 \\ & \begin{aligned} \mid(\mathbf{a} \times \mathbf{b}) & \times \mathbf{c}|=|\mathbf{a} \times \mathbf{b}|| \mathbf{c} \mid \sin \theta \\ & =3 \times 1 \times \sin 30^{\circ} \\ & =3 \times 1 \times \frac{1}{2}=\frac{3}{2} \end{aligned} \end{aligned} $

$ \begin{aligned} &(a+2 b-c)\{(a-b) \times(a-b-c)\} \\ &=(a+2 b-c)[a \times a-a \times b-a \times c-b\times a+b \times b+b \times c] \end{aligned} $

$ \begin{aligned} & =(\mathbf{a}+2 \mathbf{b}-\mathbf{c}) [-\mathbf{a} \times \mathbf{b}-\mathbf{a} \times \mathbf{c}-\mathbf{b} \times \mathbf{a}+\mathbf{b} \times \mathbf{c}] \\ & =(\mathbf{a}+2 \mathbf{b}-\mathbf{c})[\mathbf{b} \times \mathbf{a}+\mathbf{c} \times \mathbf{a}+\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}] \end{aligned} $

$ \begin{aligned} = & \mathbf{a} \cdot(\mathbf{b} \times \mathbf{a})+\mathbf{a} \cdot(\mathbf{c} \times \mathbf{a})+\mathbf{a} \cdot(\mathbf{a} \times \mathbf{b}) \\ & +\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c}) \\ & +2 \mathbf{b} \cdot(\mathbf{b} \times \mathbf{a})+2 \mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})+2 \mathbf{b}(\mathbf{a} \times \mathbf{b}) \\ & +2 \mathbf{b} \cdot(\mathbf{b} \times \mathbf{c}) \\ & -\mathbf{c} \cdot(\mathbf{b} \times \mathbf{a})-\mathbf{c}(\mathbf{c} \times \mathbf{a})-\mathbf{c}(\mathbf{a} \times \mathbf{b})-\mathbf{c}(\mathbf{b} \times \mathbf{c}) \\ = & {[\mathbf{a b c}]+2[\mathbf{b c a}]-[\mathbf{c b a}]-[\mathbf{c a b}] } \\ = & {[\mathbf{a b c}]+2[\mathbf{a b c}]+[\mathbf{a b c}]-[\mathbf{a b c}]=3[\mathbf{a b c}] } \end{aligned} $

Given that

$ \begin{aligned} \mathrm{OP} & =\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathrm{OQ}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\ \mathbf{a} & =\hat{\mathbf{i}}+\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ l_1 & =\mathbf{r}_1=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ l_2 & =\mathbf{r}_2=-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}+\mu(\hat{\mathbf{j}}-\hat{\mathbf{k}}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Let point of intersection of first line.

$ \begin{aligned} &\begin{aligned} \mathbf{L} & =\mathbf{P}+\lambda \mathbf{a}=(1+\lambda) \hat{\mathbf{i}}+(-1+\lambda) \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ M & =\mathbf{Q}+\mu \mathbf{b}=-\hat{\mathbf{i}}+(1+\mu) \hat{\mathbf{j}}+(1-\mu) \hat{k} \end{aligned}\\ &\text { Line passing through } L \text { and } M\\ &\begin{array}{lc} \therefore \mathbf{L M} \| \mathbf{C} \\ \mathbf{L M}=(-2-\lambda) \hat{\mathbf{i}}+(2+\mu-\lambda) \hat{\mathbf{j}}+(2-\mu) \hat{k} \end{array} \end{aligned} $

$ \begin{aligned} & \mathbf{L} \mathbf{M}=t(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & -2-\lambda=t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & \Rightarrow \quad \lambda=-2-t \\ & 2+\mu-\lambda=-t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ & 2-\mu=t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\\ & \Rightarrow \quad \mu=2-t \\ & t=-6, \lambda=4 \\ & \mu=8 \\ & \mathbf{M}=-\hat{\mathbf{i}}+9 \hat{\mathbf{j}}-7 \hat{\mathbf{k}} \\ & \mathbf{P M}=-2 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} & \mathbf{p}=(a+1) \hat{\mathbf{i}}+a \hat{\mathbf{j}}+a \hat{\mathbf{k}} \\ & \mathbf{q}=a \hat{\mathbf{i}}+(a+1) \hat{\mathbf{j}}+a \hat{\mathbf{k}} \\ & \mathbf{r}=a \hat{\mathbf{i}}+a \hat{\mathbf{j}}+(a+1) \hat{\mathbf{k}} \end{aligned} $

$\mathbf{p}, \mathbf{q}$ and $\mathbf{r}$ are coplanar

$ \begin{aligned} & \left|\begin{array}{ccc} a+1 & a & a \\ a & a+1 & a \\ a & a & a+1 \end{array}\right|=0 \\ \Rightarrow & (3 a+1)\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & a+1 & a \\ a & a & a+1 \end{array}\right|=0 \\ \Rightarrow & (3 a+1)\left[(a+1)^2-a^2-1\left(a^2+a-a^2\right)+1\right. \\ & \left.\left(a^2-a^2-a\right)\right]=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow(3 a+1)[2 a+1-a-a] \Rightarrow a=-\frac{1}{3} \\ & \mathbf{p} \cdot \mathbf{q}=-\frac{2}{9}-\frac{2}{9}+\frac{1}{9}=\frac{-3}{9}=\frac{-1}{3} \\ & \begin{aligned} & \Rightarrow|\mathbf{r} \times \mathbf{q}|=\sqrt{\left(\frac{4}{9}-\frac{1}{9}\right)^2-\left(\frac{-2}{9}-\frac{1}{9}\right)^2} +\left(\frac{1}{9}+\frac{2}{9}\right)^2 \end{aligned} \\ & \\ & =\sqrt{\frac{9-9+9}{81}}=\frac{1}{3} \\ & 3(\mathbf{p} \cdot \mathbf{q})^2-\lambda|\mathbf{r} \times \mathbf{q}|^2=0 \Rightarrow \lambda=1 \end{aligned} $

$ \text { } \begin{aligned} \mathbf{a}= & \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\ & \mathbf{b}=-2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & \mathbf{c}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\ \Rightarrow & (\mathbf{b} \times \mathbf{c}) \times \mathbf{a}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ \Rightarrow & (\mathbf{b} \cdot \mathbf{a}) \mathbf{c}-(\mathbf{c} \cdot \mathbf{a}) \mathbf{b}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ \Rightarrow & (-2+20+8) \mathbf{c}-(3-8+16) \mathbf{b} \\ & =x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ \Rightarrow & 100 \hat{\mathbf{i}}+107 \hat{\mathbf{j}}-82 \hat{\mathbf{k}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ \Rightarrow & x+y-z=100-107+82=75 \end{aligned} $

$ \begin{aligned} & \mathbf{A B}=5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+15 \hat{\mathbf{k}} \\ & \mathbf{A C}=7 \hat{\mathbf{i}}+20 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} \cos \theta & =\frac{35-80+45}{|\mathbf{A B}||\mathbf{A C}|}=0 \\ \theta & =90^{\circ} \end{aligned} $

$ \text { We have, } \mathbf{x} \cdot(\mathbf{c}-\mathbf{b})=\mathbf{a} \cdot \mathbf{c}-\mathbf{a} \cdot \mathbf{b} $

$ \begin{aligned} & \quad=\mathbf{a}(\mathbf{c}-\mathbf{b}) \\ & (\mathbf{x}-\mathbf{a}) \cdot(\mathbf{c}-\mathbf{b})=0 \\ & \quad \mathbf{A P} \cdot \mathbf{B C}=0 \\ & \quad \mathbf{A P} \perp \mathbf{B C} \\ & \text { and } \mathbf{x} \cdot(\mathbf{a}-\mathbf{c})=\mathbf{a} \cdot \mathbf{b}-\mathbf{b} \cdot \mathbf{c}=(\mathbf{a}-\mathbf{c}) \mathbf{b} \\ & \Rightarrow(\mathbf{x}-\mathbf{b}) \cdot(\mathbf{a}-\mathbf{c})=0 \\ & \quad \mathbf{B P} \cdot \mathbf{A C}=0 \\ & \therefore \quad \mathbf{B P} \perp \mathbf{A C} \end{aligned} $

$\because$ Point of intersection of altitude is called orthocentre.

$\because P$ is orthocentre of $\triangle A B C$.

Given $|\mathbf{a}|=2,|\mathbf{b}|=3,|\mathbf{c}|=5$

$ \begin{aligned} & (a \cdot b)=\frac{\pi}{3} \\ & a \cdot b=3 \text { and } b \cdot c=\frac{15}{2} \end{aligned} $

Given, $|\mathbf{a}+\mathbf{b}+\mathbf{c}|=\sqrt{69}$

On squaring both sides, we get

$ \begin{aligned} & |\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) =69\\ & \Rightarrow \quad 4+9+25+2\left(3+\frac{15}{2}+\mathbf{c} \cdot \mathbf{a}\right)=69 \\ & \Rightarrow 2\left(\frac{21}{2}+\mathbf{c} \cdot \mathbf{a}\right)=31 \\ & \Rightarrow \quad 21+2 \cdot \mathbf{c}=31 \\ & \Rightarrow \quad 2(\mathbf{c} \cdot \mathbf{a})=10 \\ & \Rightarrow \quad \mathbf{c} \cdot \mathbf{a}=5 \\ & \Rightarrow \quad|\mathbf{c}||\mathbf{a}| \cos \theta=5 \\ & \Rightarrow \quad 5 \times 2 \cos \theta=5 \\ & \Rightarrow \quad \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \end{aligned} $

$ \begin{aligned} &\text { We have, } \mathbf{O A}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \text {, }\\ &\begin{aligned} & \mathrm{OB}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \mathrm{OC}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} \end{aligned} $

$ \mathrm{OD}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} $

Let $\mathbf{n}_1$ is the vector normal to the face $A B C$

$ \because \mathbf{n}_1=\mathbf{A B} \times \mathbf{A C} $

$ \begin{aligned} \mathbf{n}_1 & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & -2 & 3 \\ 0 & -3 & 2 \end{array}\right| \\ & =\hat{\mathbf{i}}(-4+9)+\hat{\mathbf{j}}(0)+\hat{\mathbf{k}}(0)=5 \hat{\mathbf{i}} \end{aligned} $

Let $\mathbf{n}_2$ be the vector normal to face $A B D$

$ \because \mathbf{n}_2=\mathbf{A B} \times \mathbf{A D} $

$ n_2=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & -2 & 3 \\ 1 & 0 & 2 \end{array}\right|=\hat{\mathbf{i}}(-4)-\hat{\mathbf{j}}(-3)+\hat{\mathbf{k}}(2) $

$ \mathbf{n}_2=-4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} $

Now, angle between $\mathbf{n}_1$ and $\mathbf{n}_2$

$ \begin{aligned} \cos \theta & =\frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{\left|\mathbf{n}_1\right|\left|\mathbf{n}_2\right|}=\frac{-20}{5 \sqrt{29}} \\ \theta & =\cos ^{-1}\left(\frac{-4}{\sqrt{29}}\right) \end{aligned} $

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

$ \begin{aligned} & \mathbf{a} \cdot \mathbf{b}=0 \\ & (\mathbf{a}-\mathbf{c}) \cdot(\mathbf{b}+\mathbf{c})=0 \\ & \mathbf{a} \cdot \mathbf{b}+\mathbf{a} \cdot \mathbf{c}-\mathbf{c} \cdot \mathbf{b}-|\mathbf{c}|^2=0 \\ & \mathbf{a} \cdot \mathbf{c}-\mathbf{b} \cdot \mathbf{c}=1 \\ & \text { Also, } \mathbf{c}=l \mathbf{c}+m \mathbf{b}+n(\mathbf{a} \times \mathbf{b}) \\ & \mathbf{a} \cdot \mathbf{c}=l(\mathbf{a})^2+m(\mathbf{a} \cdot \mathbf{b})+n(\mathbf{a} \cdot(\mathbf{a} \times \mathbf{b}) \\ & \mathbf{a} \cdot \mathbf{c}=l \end{aligned} $

Similarly, $\mathbf{b} \cdot \mathbf{c}=m$

$ \begin{aligned} & \begin{array}{l} \therefore l-m=1 \Rightarrow l^2+m^2-2 l m=1 \\ \text { Now, }(\mathbf{c}-l \mathbf{a}-m \mathbf{b})^2=(n(\mathbf{a} \times \mathbf{b}))^2 \\ \Rightarrow \mathbf{c}^2+l^2 \mathbf{a}^2+m^2 \mathbf{b}-2 m \mathbf{b} \cdot \mathbf{c}+2 l m \mathbf{a} \cdot \mathbf{b}-2 l \mathbf{a} \cdot \mathbf{c}=n^2|\mathbf{a}|^2|\mathbf{b}|^2 \\ \Rightarrow 1+l^2+m^2-2 m^2-2 l^2=n^2 \\ \Rightarrow 1-l^2-m^2=n^2 \\ \Rightarrow n^2=-2 l m \quad\left(\because-2 l m=1-l^2-m^2\right) \end{array} \end{aligned} $

Let $\mathbf{n}$ is the normal to the face $O A B$

$ \because \quad \mathbf{n}=\mathbf{O A} \times \mathbf{O B} $

$ \begin{aligned} & \mathbf{n}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{array}\right| \\ & \mathbf{n}=\hat{\mathbf{i}}(5)-\hat{\mathbf{j}}(\mathbf{l})+\hat{\mathbf{k}}(-3) \\ & \mathbf{n}=5 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}} \end{aligned} $

Direction ratio of edge $\mathbf{B C}=-3,0,-1$

Now, angle between $\mathbf{n}$ and edge $\mathbf{B C}$ is

$ \begin{aligned} && \sin \theta & =\left|\frac{-15-0+3}{\sqrt{35} \sqrt{10}}\right|=\frac{-12}{\sqrt{350}} \times \frac{\sqrt{2}}{\sqrt{2}} \\ \therefore & & \sin \theta & =\frac{6 \sqrt{2}}{5 \sqrt{7}} \\ & \Rightarrow & \theta & =\sin ^{-1}\left(\frac{6 \sqrt{2}}{5 \sqrt{7}}\right) \end{aligned} $

$ \begin{aligned} \mathbf{A B} & =-3 \hat{\mathbf{i}}-3 \hat{\mathbf{k}} \\ \mathbf{B C} & =4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\ \mathbf{C A} & =-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+7 \hat{\mathbf{k}} \end{aligned} $

Given $\alpha, \beta$ and $\gamma$ are angles between the sides

$ \begin{aligned} \because \cos \alpha= & \frac{\mathbf{A B} \cdot \mathbf{B C}}{|\mathbf{A B}||\mathbf{B C}|}=0 \\ \cos \beta= & \frac{\mathbf{B C} \cdot \mathbf{C A}}{|\mathbf{B C}||\mathbf{C A}|}=\frac{-36}{6 \sqrt{54}}=-\frac{2}{\sqrt{6}} \\ \cos \gamma= & \frac{\mathbf{C A} \cdot \mathbf{A B}}{|\mathbf{C A}||\mathbf{A B}|}=\frac{-18}{6 \sqrt{54}}=\frac{2}{\sqrt{12}} \\ \sin ^2 \alpha+ & \sin ^2 \beta+\sin ^2 \gamma=\sqrt{\left(1-\cos ^2 \alpha\right)} \\ & \quad+\left(\sqrt{1-\cos ^2 \beta}\right)+\left(\sqrt{1-\cos ^2 \gamma}\right) \\ = & 1+\frac{2}{6}+\frac{8}{12}=1+1=2 \end{aligned} $

We have, $A(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}), B(5 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}})$ and $C(-13 \hat{\mathbf{i}}-11 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})$,

$ \begin{aligned} \mathbf{A B} & =\lambda \mathbf{B C} \\ 3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} & =\lambda(-18 \hat{\mathbf{i}}-12 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \\ \because-18 \lambda & =3 \\ \lambda & =-\frac{3}{18} \end{aligned} $

Given, $\mathbf{A C}=\mu \mathbf{C B}$

$ \begin{aligned} -15 \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+5 \hat{\mathbf{k}} & =\mu(18 \hat{\mathbf{i}}+12 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}) & \\ \mu & =-\frac{15}{18} & \\ \therefore \,\,& \lambda+\mu =-\frac{3}{18}-\frac{15}{18}=-\frac{18}{18}=-1 \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} & A C=O C-O A \\ & O C=O A+A C \\ & O C=a+3 A B \\ & O C=a+3(b-a) \\ & O C=3 b-2 a \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned}Similarly,\,\, \mathbf{B D} & =\mathbf{O D}-\mathbf{O B} \\ \mathbf{O D} & =\mathbf{O B}+\mathbf{B D} \\ & =\mathbf{b}+2 \mathbf{B A} \\ & =\mathbf{b}+2(\mathbf{a}-\mathbf{b}) \\ \mathbf{O D} & =2 \mathbf{a}-\mathbf{b} \end{aligned}\\ &\begin{aligned} C D & =O D-O C \\ & =(2 a-b)-(3 b-2 a) \\ C D & =4 a-4 b \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & |\mathbf{a}-\mathbf{b}|^2+|\mathbf{b}-\mathbf{c}|^2+|\mathbf{c}-\mathbf{a}|^2=15 \\ \Rightarrow \quad & |\mathbf{a}|^2+|\mathbf{b}|^2-2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^2+|\mathbf{c}|^2 \\ & -2 \mathbf{b} \cdot \mathbf{c}+|\mathbf{c}|^2+|\mathbf{a}|^2-2 \mathbf{c} \cdot \mathbf{a}=15 \\ \Rightarrow \quad & 6-2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})=15 \\ \Rightarrow \quad & \mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}=-\frac{9}{2} \\ \quad & |\mathbf{a}-\mathbf{b}-\mathbf{c}|^2-4(\mathbf{b} \cdot \mathbf{c}) \\ \Rightarrow \quad & |\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2 \\ & +2(-\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}-\mathbf{c} \cdot \mathbf{a})-4(\mathbf{b} \cdot \mathbf{c}) \\ & =3-2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \\ & =3-2\left(-\frac{9}{2}\right) \\ & =3+9=12 \end{aligned} \end{aligned} $

$\mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & p & -3 \\ p & -3 & 1\end{array}\right|$

$ \begin{aligned} \hat{\mathbf{i}}(p-9)-\hat{\mathbf{j}}(1+3 p)+k\left(-3-p^2\right) & \\ |\mathbf{a} \times \mathbf{b}|^2 & =(p-9)^2+(3 p+1)^2+\left(p^2+3\right)^2 \\ \mathbf{a} \times \mathbf{c} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & p & -3 \\ -3 & 1 & 2 \end{array}\right| \\ & =\hat{\mathbf{i}}(2 p+3)-\hat{\mathbf{j}}(2-9)+\hat{\mathbf{k}}(1+3 p) \\ |\mathbf{a} \times \mathbf{c}|^2 & =(2 p+3)^2+49+(3 p+1)^2 \end{aligned} $

Given, $|\mathbf{a} \times \mathbf{b}|=|\mathbf{a} \times \mathbf{c}|$

$ \begin{array}{r} \because \quad(p-9)^2+(3 p+1)^2+\left(p^2+3\right)^2=(2 p+3)^2 +49+(3 p+1)^2 \\ \Rightarrow \quad p^2+81-18 p+\left(p^2+3+2 p+3\right) \left(p^2+3-2 p-3\right)=49 \\ \Rightarrow \quad p^2+81-18 p+\left(p^2+2 p+6\right)\left(p^2-2 p\right) =49 \\ \Rightarrow p^2+81-18 p+\left(p^2+2 p+6\right)\left(p^2-2 p\right) =49 \end{array} $

$ \begin{aligned} &\Rightarrow p^4+3 p^2-30 p+32=0\\ &\text { Now put } p=2.16+12-.60+32=0\\ &\because \quad p=2 \end{aligned} $

$(a \times b) \times(c \times d)$

$ \begin{aligned} & =[\mathbf{a} \mathbf{b} \mathbf{d}] \mathbf{c}-[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{d} \\ {[\mathbf{a} \mathbf{b} \mathbf{d}] \mathbf{c} } & =8(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ & =24 \hat{\mathbf{i}}-8 \hat{\mathbf{j}}+16 \hat{\mathbf{k}} \\ {[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{d} } & =-7(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & =-7 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-7 \hat{\mathbf{k}} \end{aligned} $

$[\mathbf{a} \mathbf{b} \mathbf{d}] \mathbf{c}-[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{d}=31 \hat{\mathbf{i}}-\hat{\mathbf{j}}+23 \hat{\mathbf{k}}$

Given the vectors $ a \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}, \hat{\mathbf{i}} + b \hat{\mathbf{j}} + \hat{\mathbf{k}} $, and $ \hat{\mathbf{i}} + \hat{\mathbf{j}} + c \hat{\mathbf{k}} $, we need to verify that they are coplanar. When vectors are coplanar, it means they can be expressed as a linear combination of the other vectors.

Thus, we can write:

$ a \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}} = x (\hat{\mathbf{i}} + b \hat{\mathbf{j}} + \hat{\mathbf{k}}) + y (\hat{\mathbf{i}} + \hat{\mathbf{j}} + c \hat{\mathbf{k}}) $

where $ x $ and $ y $ are scalars, and not both zero.

Expanding both sides, we get:

$ a \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}} = (x+y) \hat{\mathbf{i}} + (bx + y) \hat{\mathbf{j}} + (x + cy) \hat{\mathbf{k}} $

Equating the coefficients of $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$, we derive:

$ a = x + y \tag{1} $

$ 1 = bx + y \tag{2} $

$ 1 = x + cy \tag{3} $

Now, from Equation (1), we have:

$ 1-a = 1 - (x + y) $

Substitute the rearranged form of Equation (2):

$ 1-b = 1 - (bx + y) = \frac{x-1+y}{x} $

Substitute from Equation (3):

$ 1-c = 1 - (x + cy) = \frac{y-1+x}{y} $

Calculate the desired expression:

$ \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = \frac{1}{1-(x+y)} + \frac{x}{x-1+y} + \frac{y}{y-1+x} $

By simplifying:

$ = \frac{1-x-y}{1-(x+y)} = 1 $

Thus, the expression simplifies to $1$.

Given, $\mathbf{A B}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}$

$ \mathrm{BC}=6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} $

We know that $\mathbf{A B}+\mathbf{B C}+\mathbf{C A}=0$

$ \begin{array}{ll} \Rightarrow & (8 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})=\mathbf{A C} \\ \therefore & |\mathbf{A B}|=\sqrt{4+9+36}=7 \\ & |\mathbf{B C}|=\sqrt{36+4+9}=7 \\ \text { and } & |\mathbf{A C}|=\sqrt{64+1+9}=\sqrt{74} \end{array} $

$\therefore$ Perimeter of $\triangle A B C$ is

$ 7+7+\sqrt{74}=14+\sqrt{74} $

To find the orthogonal projection of vector $\mathbf{a} = 2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}$ on vector $\mathbf{b} = \hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + \hat{\mathbf{k}}$, we use the formula for the projection of $\mathbf{a}$ onto $\mathbf{b}$:

$ \text{Projection}_{\mathbf{b}}(\mathbf{a}) = \frac{(\mathbf{a} \cdot \mathbf{b}) \mathbf{b}}{|\mathbf{b}|^2} $

Step-by-step, this involves:

Calculate the dot product $\mathbf{a} \cdot \mathbf{b}$:

$ \mathbf{a} \cdot \mathbf{b} = (2)(1) + (3)(-2) + (3)(1) = 2 - 6 + 3 = -1 $

Find the magnitude squared of $\mathbf{b}$, $ |\mathbf{b}|^2 $:

$ |\mathbf{b}|^2 = (1)^2 + (-2)^2 + (1)^2 = 1 + 4 + 1 = 6 $

Substitute these values into the projection formula:

$ \text{Projection}_{\mathbf{b}}(\mathbf{a}) = \frac{-1(\hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + \hat{\mathbf{k}})}{6} = \frac{-\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - \hat{\mathbf{k}}}{6} $

Thus, the orthogonal projection vector of $\mathbf{a}$ on $\mathbf{b}$ is:

$ \frac{-\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - \hat{\mathbf{k}}}{6} $

Given vectors are:

$\mathbf{a} = -4 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + 4 \hat{\mathbf{k}}$

$\mathbf{b} = \sqrt{2} \hat{\mathbf{i}} - \sqrt{2} \hat{\mathbf{j}}$

First, we compute $2 \mathbf{a}$ and $\frac{\mathbf{b}}{2}$:

$ 2 \mathbf{a} = -8 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}} + 8 \hat{\mathbf{k}} $

$ \frac{\mathbf{b}}{2} = \frac{1}{\sqrt{2}} \hat{\mathbf{i}} - \frac{1}{\sqrt{2}} \hat{\mathbf{j}} $

To find the angle between $2 \mathbf{a}$ and $\frac{\mathbf{b}}{2}$, we use the dot product formula:

$ \cos \theta = \frac{2 \mathbf{a} \cdot \frac{\mathbf{b}}{2}}{|2 \mathbf{a}|\left|\frac{\mathbf{b}}{2}\right|} $

This simplifies to:

$ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} $

Calculating the dot product $\mathbf{a} \cdot \mathbf{b}$:

$ \begin{aligned} \mathbf{a} \cdot \mathbf{b} &= (-4)(\sqrt{2}) + (2)(-\sqrt{2}) + (4)(0) \\ &= -4\sqrt{2} - 2\sqrt{2} \\ &= -6\sqrt{2} \end{aligned} $

Next, we find the magnitudes of $\mathbf{a}$ and $\mathbf{b}$:

$ |\mathbf{a}| = \sqrt{(-4)^2 + 2^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 $

$ |\mathbf{b}| = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 $

Substitute these into the cosine formula:

$ \cos \theta = \frac{-6\sqrt{2}}{6 \times 2} = \frac{-6\sqrt{2}}{12} = -\frac{1}{\sqrt{2}} $

The angle $\theta$ is calculated as:

$ \theta = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) $

Since $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ corresponds to $\pi - \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$:

$ \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} = 135^{\circ} $