Vector Algebra

Let $\mathbf{r}=\mathbf{a}+\lambda \mathbf{b}$

Given that, projection of $\mathbf{r}$ on $\mathbf{c}=\frac{9}{\sqrt{6}}$

$ \begin{aligned} & \text { Since, } \quad \mathbf{r} \cdot \mathbf{c}=\frac{9}{\sqrt{6}} \\ & \Rightarrow \quad \frac{\mathbf{r} \cdot \mathbf{c}}{|\mathbf{c}|}=\frac{9}{\sqrt{6}} \\ & \Rightarrow \frac{\mathbf{a} \cdot \mathbf{c}+\lambda \mathbf{b} \cdot \mathbf{c}}{\sqrt{1+1+4}}=\frac{9}{\sqrt{6}} \\ & \Rightarrow(2-1-2)+\lambda(1+2+2)=9 \\ & \Rightarrow \quad \lambda=2 \end{aligned} $

Hence, $\mathbf{r}=\mathbf{a}+\mathbf{2} \mathbf{b}=\mathbf{4} \mathbf{i}+3 \mathbf{j}-\mathbf{j}$

$ |\mathrm{r}|=\sqrt{16+9+1}=\sqrt{26} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } O A=a, O B=b, O C=c \\ & \therefore O G=\frac{a+b+c}{3}, O G=\frac{2 \times O S-1 \times O O}{2+1} \\ & O G=\frac{2 O S}{3} \Rightarrow O S=\frac{3}{2} O G \\ & O S=\frac{a+b+c}{2} \\ & \text { and } O A+O B+O C=a+b+c \\ & =2\left(\frac{a+b+c}{2}\right)=2 O S \\ & S A+S B+S C=O A-O S+O B-O S +O C-O S \end{aligned} \end{aligned} $

$ \begin{aligned} & =\mathbf{O A}+\mathbf{O B}+\mathbf{O C}-3 \mathbf{O B} \\ & =2 \mathbf{O S}-3 \mathbf{O S}=-\mathbf{O S}=\mathbf{S O} \\ \mathbf{G A} & +\mathbf{G B}+\mathbf{G C} \\ & =\mathbf{O A}-\mathbf{O G}+\mathbf{O B}-\mathbf{O G}+\mathbf{O C}-\mathbf{O G} \\ & =(\mathbf{a}+\mathbf{b}+\mathbf{c})-3(\mathbf{O G}) \\ & =\mathbf{a}+\mathbf{b}+\mathbf{c}-3\left(\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}\right) \\ & =\mathbf{0} \end{aligned} $

$ \begin{aligned} & \mathbf{a} \cdot \mathbf{a}=|\mathbf{a}|^2=5 \\ & \mathbf{b} \cdot \mathbf{b}=|\mathbf{b}|^2=5 \\ & \mathbf{c} \cdot \mathbf{c}=|\mathbf{c}|^2=5 \end{aligned} $

$ \begin{aligned} &\text { Now, }\\ &|\mathbf{a}+\mathbf{b}-\mathbf{c}|^2+|\mathbf{b}+\mathbf{c}-\mathbf{a}|^2+|\mathbf{c}+\mathbf{a}-\mathbf{b}|^2 \end{aligned} $

$ \begin{aligned} =3\left(|\mathbf{a}|^2+|\mathbf{b}|^2\right. & \left.+|\mathbf{c}|^2\right) -2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})=50 \end{aligned} $

$ \begin{aligned} & \Rightarrow \mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}=\frac{3(5+5+5)-50}{2} \\ & \Rightarrow \mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}=\frac{-5}{2} \end{aligned} $

$\mathbf{a , b}$ and $\mathbf{c}$ are coplanar

$ \Rightarrow[\mathbf{a} \mathbf{b} \mathbf{c}]=0 $

Now, $[\mathbf{a} \mathbf{b} \mathbf{d}] \mathbf{c}-0=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$

$ \begin{aligned} \Rightarrow \quad\{(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}\} \mathbf{c} & =\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ \Rightarrow \quad|\mathbf{c}| & =\frac{3}{|(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}|} \\ & =\frac{3}{\sin \frac{\pi}{6}}=6 \end{aligned} $

Given,

$ \begin{aligned} |a| & =4 \\ |b| & =5 \\ |a-b| & =3 \\ a \cdot b & =|a||b| \cos \theta \\ a \cdot b & =4 \times 5 \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

So, $(\mathbf{a}-\mathbf{b})^2=|\mathbf{a}|^2+|\mathbf{b}|^2-2 \mathbf{a} \cdot \mathbf{b}$

$ 9=16+25-40 \cos \theta $

[From Eq. (i)]

$ \therefore \quad \cos \theta=\frac{32}{40}=\frac{4}{5} $

Now, from $\triangle A B C$,

$ \begin{aligned} & A B=\sqrt{25-16}=3 \\ & B C=4 \Rightarrow A C=5 \end{aligned} $

$ \therefore \quad \cot \theta=\frac{\text { Base }}{\text { Perpendicular }}=\frac{4}{3} $

Hence, $\cot ^2 \theta=\frac{16}{9}$

(by squaring on both sides)

$|\mathbf{a}|=3,|\mathbf{b}|=5,|\mathbf{c}|=7$

$ \begin{aligned} & \because a+b+c=0 \\ & \Rightarrow a+b=-c \\ & |a+b|=|-c| \end{aligned} $

[taking mod on both sides]

$|a+b|^2=|-c|^2$ [squaring on both sides]

$ \begin{gathered} |\mathbf{a}|^2+|\mathbf{b}|^2+2 \mathbf{a} \cdot \mathbf{b}=|\mathbf{c}|^2 \\ 9+25+2 \mathbf{a} \cdot \mathbf{b}=49 \end{gathered} $

Also, $\mathbf{a b}=3 \times 5$

Hence, $9+25+2|\mathbf{a}||\mathbf{b}| \cos \theta=49$

$ \begin{aligned} & \Rightarrow 30 \cos \theta=15 \Rightarrow \cos \theta=1 / 2 \\ & \therefore \theta=60^{\circ}=\pi / 3 \end{aligned} $

So, angle is $\pi / 3$.

Given,

$ A(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}), B(-12 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}}), $

$C(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})$ and $D(\lambda \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$ are position vector of four points.

$ \begin{aligned} & \mathbf{A B}=14 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}+6 \hat{\mathbf{k}} \\ & \mathbf{B C}=-11 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \mathrm{CD}=(-1-\lambda) \hat{\mathbf{i}}+0 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \end{aligned} $

∴ For coplanar

$ \begin{array}{rlrl} \left|\begin{array}{ccc} 14 & 0 & 6 \\ -11 & -3 & 1 \\ -(\lambda+1) & 0 & -3 \end{array}\right| & =0 \\ 14(9)+6(-3(\lambda+1)) & =0 \\ 126-18 \lambda-18 & =0 \\ 18 \lambda & =108 \\ \Rightarrow & & \lambda & =6 \end{array} $

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

We know that, The orthogonal projection of $\mathbf{a}$ on

$ \begin{aligned} & \mathbf{b}(\mathbf{x})=\frac{(\mathbf{a} \cdot \mathbf{b}) \mathbf{b}}{|\mathbf{b}|^2} \\ & \Rightarrow \mathbf{x}=\frac{[(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}})]}{\left(\sqrt{(2)^2+(-1)^2+(-2)^2}\right)^2} \\ & \quad(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}})) \end{aligned} $

$ \begin{aligned} & \Rightarrow x=\frac{(1 \times 2)+2 \times(-1)+(-2) \cdot(-2)}{9}(\hat{\mathbf{2i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ & \therefore x=\frac{4}{9}(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Orthogonal projection of $\mathbf{b}$ on

$ \begin{aligned} & \mathbf{a}(\mathbf{y})=\frac{(\mathbf{b} \cdot \mathbf{a}) \mathbf{a}}{|\mathbf{a}|^2} \\ & \quad \mathbf{y}=\frac{4}{9}(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

So, $\quad \mathbf{x}-\mathbf{y}=\frac{4}{9}(\hat{\mathbf{i}}-3 \hat{\mathbf{j}})$

$ \begin{aligned} & \text { Hence, }|x-y|=\frac{4}{9} \sqrt{(1)^2+(-3)^2} \\ & \therefore \quad|x-y|=\frac{4}{9} \sqrt{10} \end{aligned} $

Given, $\mathbf{O A}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$

$ \mathbf{O B}=2 \hat{\mathbf{i}}-\hat{\mathbf{k}} \Rightarrow \mathbf{O C}=\hat{\mathbf{j}}+2 \hat{\mathbf{k}} $

$and\,\,\, \mathrm{OD}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}} $

$ \begin{aligned} So,\,\,& \mathrm{AB}=\mathrm{OB}-\mathrm{OA}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & \mathrm{AC}=\mathrm{OC}-\mathrm{OA}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \mathrm{AD}=\mathrm{OD}-\mathrm{OA}=2 \hat{\mathbf{j}}+\hat{\mathbf{k}}(\lambda-1) \end{aligned} $

Since, $A, B, C$ and $D$ are coplanar.

$ \begin{aligned} & \because \quad\left|\begin{array}{ccc} 1 & 1 & -2 \\ -1 & 2 & 1 \\ 0 & 2 & \lambda-1 \end{array}\right|=0 \\ & \Rightarrow 1(2 \lambda-2-2)-1(-\lambda+1-0)-2(-2)=0 \\ & \Rightarrow 2 \lambda-4+\lambda-1+4=0 \\ & \Rightarrow \quad 3 \lambda=1 \\ & \Rightarrow \quad \lambda=\frac{1}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\text {} \end{aligned} $

Magnitude of vector $6 \lambda \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$

$ \begin{aligned} & =\sqrt{(6 \lambda)^2+(-3)^2+(6)^2} \\ & =\sqrt{\left(6 \times \frac{1}{3}\right)^2+9+36} \text { [from Eq. (i), } \lambda=\frac{1}{3} \text { ] } \\ & =\sqrt{4+9+36}=7 \text { units } \end{aligned} $

Clearly, the point $(\mathbf{a}-\mathbf{b}+\mathbf{c})$ lies on the line because line is passing through this point.

∴ Equation of plane is

$ \begin{array}{r} {[\mathbf{r}-(2 \mathbf{a}-\mathbf{b})][(2 \mathbf{a}-\mathbf{b})-(2 \mathbf{b}-\mathbf{c})] \times[(2 c-\mathbf{a})} -(2 \mathbf{b}-\mathbf{c})]=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{array} $

Now, we have to show that the point. $(\mathbf{a}-\mathbf{b}+\mathbf{c})$ satisfying the equation plane.

Taking LHS of Eq. (i),

$ \begin{aligned} &(\mathbf{r}-2 \mathbf{a}+\mathbf{b}) \cdot[(2 \mathbf{a}-3 \mathbf{b}+\mathbf{c}) \times(-\mathbf{a}-2 \mathbf{b}+3 \mathbf{c})] \\ &=(\mathbf{r}-2 \mathbf{a}+\mathbf{b})[-4 \mathbf{a} \times \mathbf{b}+6 \mathbf{a} \times \mathbf{c}+ \\ &+3 \mathbf{b} \times \mathbf{a}-9 \mathbf{b} \times \mathbf{c}-\mathbf{c} \times \mathbf{a}-2 \mathbf{c} \times \mathbf{b}] \\ &=(\mathbf{r}-2 \mathbf{a}+\mathbf{b})[-7 \mathbf{a} \times \mathbf{b}+7 \mathbf{a} \times \mathbf{c}-7 \mathbf{b} \times \mathbf{c}] \\ &=(\mathbf{r}-2 \mathbf{a}+\mathbf{b})(-\mathbf{a} \times \mathbf{b}+\mathbf{a} \times \mathbf{c}-\mathbf{b} \times \mathbf{c}) \end{aligned} $

On putting $\mathbf{r}=\mathbf{a}-\mathbf{b}+\mathbf{c}$,

$ \begin{aligned} & =(-\mathbf{a}+\mathbf{c})(-\mathbf{a} \times \mathbf{b}+\mathbf{a} \times \mathbf{c}-\mathbf{b} \times \mathbf{c}) \\ & \Rightarrow[\mathrm{a}: \mathbf{b} \mathbf{c}]-[\mathrm{c} \mathbf{a} \mathbf{b}]=0 \end{aligned} $

Hence, the point of intersection of line and plane is $\mathbf{a}-\mathbf{b}+\mathbf{c}$.

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

$ \begin{aligned} &\,\,\,\,\,\,\,\,\,\,\, \mathbf{b}=6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\ &and\,\,\, \mathbf{c}=3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-12 \hat{\mathbf{k}} \end{aligned} $

The projection of $\mathbf{b}$ on $\mathbf{a}=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2} \mathbf{a}$

$ \begin{aligned} &\Rightarrow \mathbf{p}=\frac{6-4-6}{\sqrt{\left(1^2+(-2)^2+2^2\right)^2}} \mathbf{a} \Rightarrow \mathbf{p}=-\frac{4}{9} \mathbf{a}\\ &\text { The projection of } \mathbf{c} \text { on } \mathbf{a}=\frac{\mathbf{a} \cdot \mathbf{c}}{|\mathbf{a}|^2} \mathbf{a}\\ &\begin{array}{ll} \Rightarrow & \mathbf{q}=\frac{3+8-24}{\sqrt{\left(1^2+(-2)^2+2^2\right)^2}} \mathbf{a} \\ \Rightarrow & \mathbf{q}=-\frac{13}{9} \mathbf{a} \\ \therefore & 13 \mathbf{p}=13\left(-\frac{4}{9} \mathbf{a}\right)=4\left(-\frac{13}{9} \mathbf{a}\right)=4 \mathbf{q} \end{array} \end{aligned} $

Given, $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$

$ \begin{aligned} & \mathbf{b}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}} \\ & \mathbf{c}=\hat{\mathbf{i}}-4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \end{aligned} $

Since, $\mathbf{r}$ is perpendicular to both $\mathbf{b}$ and $\mathbf{c}$.

$ \begin{aligned} & \therefore \mathbf{r}=t\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & -1 & 5 \\ 1 & -4 & -2 \end{array}\right| \\ & =[\hat{\mathbf{i}}(2+20)-\hat{\mathbf{j}}(-6-5)+\hat{\mathbf{k}}(-12+1)] t \\ & =(22 \hat{\mathbf{i}}+11 \hat{\mathbf{j}}-11 \hat{\mathbf{k}}) t=22 t \hat{\mathbf{i}}+11 \hat{\mathbf{j}}-11 \hat{\mathbf{k}} \\ & \because \mathbf{r} \cdot \mathbf{a}=11 \\ & \therefore(22 t \hat{\mathbf{i}}+11 t \hat{\mathbf{j}}-11 t \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})=11 \end{aligned} $

$ \begin{array}{lrl} \Rightarrow & 22 t+22 t-33 t & =11 \\ \Rightarrow & 11 t & =11 \\ \Rightarrow & t & =1 \end{array} $

Thus, $\mathbf{r}=22 \hat{\mathbf{i}}+11 \hat{\mathbf{j}}-11 \hat{\mathbf{k}}$

Since, $\mathbf{r} \cdot(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=22-11-11=0$

$\therefore \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ is perpendicular to $\mathbf{r}$.

$ \begin{aligned} &\text { Volume of a tetrahedron }\\ &=\frac{1}{6}\left[\begin{array}{lll} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{array}\right] \Rightarrow 2=\frac{1}{6}\left[\begin{array}{ccc} 1 & -\lambda & 1 \\ \lambda & -1 & -1 \\ 1 & 1 & \lambda \end{array}\right] \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 2=\frac{1}{6}\left[1(-\lambda+1)+\lambda\left(\lambda^2+1\right)+1(\lambda+1)\right] \\ & \Rightarrow \quad 12=-\lambda+1+\lambda^3+\lambda+\lambda+1 \\ & \quad 12=\lambda^3+\lambda+2 \\ & \Rightarrow \quad \lambda^3+\lambda-10=0 \\ & \quad \therefore \quad \lambda=2 \\ & \text { Hence, }|\lambda \hat{\mathbf{i}}-3 \lambda \hat{\mathbf{j}}+3 \hat{\mathbf{k}}| \\ & \begin{aligned} =|2 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}| & =\sqrt{(2)^2+(-6)^2+(3)^2} \\ & =\sqrt{4+36+9}=\sqrt{49}=7 \end{aligned} \end{aligned} $

$ \begin{aligned} & \mathrm{OA}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}=\mathbf{a} \\ & \mathrm{OB}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}=\mathbf{b} \\ & \mathrm{OC}=4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}=\mathbf{c} \end{aligned} $

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

$ \begin{aligned} \mathbf{P C} & =\mathbf{O C}-\mathbf{O P} \\ & =(4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})-(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ \mathbf{P C} & =3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}} \end{aligned} $

Here, $l=3, m=-4, n=6$

Then, $l+3 m+2 n=3-12+12=3$

$ \begin{aligned} & B C=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & C D=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \end{aligned} $

Diagonal $\mathbf{B D}=\mathbf{B C}+\mathbf{C D}$

$ \begin{aligned} & =(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ & =3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} $

Diagonal $\mathbf{A C}=\mathbf{A B}+\mathbf{B C}$

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

Angle between $\mathbf{A C}$ and $\mathbf{B D}$

$ \begin{aligned} \cos \theta & =\frac{\mathbf{A C} \cdot \mathbf{B D}}{|\mathbf{A C}||\mathbf{B D}|}=\frac{-3}{\sqrt{19} \sqrt{11}}=\frac{-3}{\sqrt{209}} \\ \Rightarrow \tan \theta & =\frac{-10 \sqrt{2}}{3} \end{aligned} $

$ \begin{aligned} & \text { } \begin{aligned} \mathbf{a} & =\lambda \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \\ \mathbf{b} & =3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}} \end{aligned} \\ & \text { and } \quad \mathbf{c}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}} \end{aligned} $

Volume of parallelopiped = 61

$ \begin{aligned} & {[\mathbf{a b c}]=61 } \\ \Rightarrow & \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=61 \\ \Rightarrow & \left|\begin{array}{ccc} \lambda & 3 & 4 \\ 3 & -1 & \lambda \\ \lambda & 1 & -3 \end{array}\right|=61 \\ \Rightarrow & \lambda(3-\lambda)-3\left(-9-\lambda^2\right)+4(3+\lambda)=61 \\ \Rightarrow & 3 \lambda-\lambda^2+27+3 \lambda^2+12+4 \lambda-61=0 \\ \Rightarrow & 2 \lambda^2+7 \lambda-22=0 \\ \Rightarrow & 2 \lambda^2+11 \lambda-4 \lambda-22=0 \\ \Rightarrow & 2 \lambda(\lambda+11)-4(\lambda+11)=0 \\ \Rightarrow & \lambda=2,-11 \end{aligned} $

$ \begin{aligned} & \text { Given, } \mathbf{a} \cdot \mathbf{b}=0 \text { and }|\mathbf{a}|=8,|\mathbf{b}|=3 \\ & \text { } \begin{aligned}Now, |\mathbf{a}-2 \mathbf{b}|^2 & =(\mathbf{a}-2 \mathbf{b}) \cdot(\mathbf{a}-2 \mathbf{b}) \\ & =|\mathbf{a}|^2+4|\mathbf{b}|^2 \end{aligned} \end{aligned} $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because \mathbf{a} \cdot \mathbf{b}=0) $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=(8)^2+4(3)^2 $

$ \begin{aligned} &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =64+4 \times 9=64+36=100 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \therefore \quad|a-2 b|=\sqrt{100}=10 \end{aligned} $

Given, $\mathbf{a}$ and $\mathbf{b}$ be non-collinear vectors.

If the vectors $(\lambda-1) \mathbf{a}+2 \mathbf{b}$ and $3 \mathbf{a}+\lambda \mathbf{b}$ are collinear.

Now, $(\lambda-1) \mathbf{a}+2 \mathbf{b}=P(3 \mathbf{a}+\lambda \mathbf{b})$

$ \Rightarrow\{(\lambda-1)-3 P\} \mathbf{a}+\{2-\lambda P\} \mathbf{b}=0 $

Since, $\mathbf{a}$ and $\mathbf{b}$ are non-collinear,

So, $\quad \lambda-1-3 P=0$ and $2-\lambda P=0$

$ \begin{aligned} & \Rightarrow \quad \lambda-1=3 \cdot P \text { and } P=\frac{2}{\lambda} \\ & \Rightarrow \quad \lambda-1 \Rightarrow \frac{6}{\lambda}=\lambda^2-\lambda-6=0 \\ & \Rightarrow \quad \lambda=-2 \text { or } 3 \end{aligned} $

Thus, the set of all possible value of $\lambda$ is $\{-2,3\}$.

Vectors, $\mathbf{p}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$, $\mathbf{q}=d \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \mathbf{r}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ forming a $\triangle A B C$ are such that, $\mathbf{p}=\mathbf{q}+\mathbf{r}$.

where, $a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$

$ \begin{aligned} & =(d \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})+(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ & \Rightarrow a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}=(d+3) \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned} $

Comparing coefficient of $\hat{\mathbf{i}}, \hat{\mathbf{j}}$ and $\hat{\mathbf{k}}$, we get

$ a=d+3, b=4 \text { and } c=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Now, area of $\triangle A B C$ is $5 \sqrt{6}$ sq units

$ \Rightarrow \quad \frac{1}{2}|\mathbf{q} \times \mathbf{r}|=5 \sqrt{6} $

$ \begin{aligned} & \Rightarrow \frac{1}{2}\left\|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ d & 3 & 4 \\ 3 & 1 & -2 \end{array}\right\|=5 \sqrt{6} \\ & \Rightarrow \frac{1}{2} \sqrt{(-10)^2+(2 d+12)^2+(d-9)^2}=5 \sqrt{6} \\ & \Rightarrow \quad \frac{1}{4}\left(5 d^2+30 d+325\right)=150 \\ & \Rightarrow \quad \quad 5 d^2+30 d+325=600 \\ & \Rightarrow \quad \quad d^2+6 d-55=0 \\ & \therefore d=-11,5 \quad a=-8,8 \quad \text { [from Eq. (i)] } \\ & \quad b=4, \text { and } c=2 \quad|a|+|b|+|c| \quad \text { Now, } \\ & =| \pm 8|+|4|+|2|=14 \end{aligned} $

we have

$ \begin{aligned} & (\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}+(\mathbf{a} \cdot \mathbf{b}) \mathbf{b} \\ & =(4-2 \beta-\sin \alpha) \mathbf{b}+\left(\beta^2-1\right) \mathbf{c} \end{aligned} $

Comparing on both side, we get

$ \begin{aligned} \mathbf{a} \cdot \mathbf{c}+\mathbf{a} \cdot \mathbf{b} & =4-2 \beta-\sin \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \text { and } \quad-\mathbf{a} \cdot \mathbf{b} & =\left(\beta^2-1\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Given, $(\mathbf{c} \cdot \mathbf{c}) \mathbf{a}=\mathbf{c}$

$ \begin{array}{lc} \Rightarrow & (\mathbf{c} \cdot \mathbf{c})(\mathbf{a} \cdot \mathbf{c})=(\mathbf{c} \cdot \mathbf{c}) \\ \Rightarrow & \mathbf{a} \cdot \mathbf{c}=1 \end{array} $

Now, from Eq. (i),

$ 1+\mathbf{a} \cdot \mathbf{b}=4-2 \beta-\sin \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Using Eq. (ii), we get

$ \begin{aligned} & 1+1-\beta^2=4-2 \beta-\sin \alpha \\ \Rightarrow & \beta^2-2 \beta+2=\sin \alpha \\ \Rightarrow & (\beta-1)^2+1=\sin \alpha \end{aligned} $

It is possible only when $\beta=1$ and $\sin \alpha=1$

$ \Rightarrow \quad \beta=1 \text { and } \alpha=\frac{\pi}{2} $

Now, $\sin (\alpha+\beta)=\sin \left(\frac{\pi}{2}+1\right) \equiv \cos (1)$

Given the position vectors of $\mathbf{P}$ and $\mathbf{Q}$ are:

$ \begin{aligned} \mathbf{P} &= \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - 7 \hat{\mathbf{k}} \\ \mathbf{Q} &= 5\hat{\mathbf{i}} - 3\hat{\mathbf{j}} + 4\hat{\mathbf{k}} \end{aligned} $

To find the vector $\mathbf{PQ}$, we calculate:

$ \mathbf{PQ} = \mathbf{Q} - \mathbf{P} = (5\hat{\mathbf{i}} - 3\hat{\mathbf{j}} + 4\hat{\mathbf{k}}) - (\hat{\mathbf{i}} + 2\hat{\mathbf{j}} - 7\hat{\mathbf{k}}) $

Simplifying the components, we get:

$ \mathbf{PQ} = (5 - 1)\hat{\mathbf{i}} + (-3 - 2)\hat{\mathbf{j}} + (4 + 7)\hat{\mathbf{k}} = 4\hat{\mathbf{i}} - 5\hat{\mathbf{j}} + 11\hat{\mathbf{k}} $

The direction ratios of $\mathbf{PQ}$ are $a = 4$, $b = -5$, and $c = 11$.

To find the cosine of the angle $\gamma$ between the vector $\mathbf{PQ}$ and the $Z$-axis, we use the relation:

$ \cos \gamma = \frac{c}{\sqrt{a^2 + b^2 + c^2}} $

Substituting the values, we have:

$ \cos \gamma = \frac{11}{\sqrt{4^2 + (-5)^2 + 11^2}} = \frac{11}{\sqrt{16 + 25 + 121}} = \frac{11}{\sqrt{162}} $

To solve this problem, we need to evaluate the given vector conditions.

We are given that $|\mathbf{a}+\mathbf{b}+\mathbf{c}| = 1$, where $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are unit vectors, meaning that $|\mathbf{a}|^2 = |\mathbf{b}|^2 = |\mathbf{c}|^2 = 1$. Additionally, $\mathbf{a}$ is perpendicular to $\mathbf{b}$, which implies $\mathbf{a} \cdot \mathbf{b} = 0$.

To proceed, we expand and square the magnitude of the vector sum:

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

$ | \mathbf{a} |^2 + | \mathbf{b} |^2 + | \mathbf{c} |^2 + 2(\mathbf{a} \cdot \mathbf{b}) + 2(\mathbf{b} \cdot \mathbf{c}) + 2(\mathbf{c} \cdot \mathbf{a}) = 1 $

Substituting the known values:

$ 1 + 1 + 1 + 0 + 2(\mathbf{b} \cdot \mathbf{c}) + 2(\mathbf{c} \cdot \mathbf{a}) = 1 $

This simplifies to:

$ 3 + 2 \cos \beta + 2 \cos \alpha = 1 $

Therefore:

$ 2 \cos \alpha + 2 \cos \beta = -2 $

Dividing the entire equation by 2 gives:

$ \cos \alpha + \cos \beta = -1 $

To determine the equation of the line through a given point and parallel to a vector $\mathbf{a}$, consider the following:

Given:

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

$\mathbf{a} \cdot \hat{\mathbf{i}} = 1$

First, find the vector $\mathbf{a}$. Assume $\mathbf{a}$ can be expressed as $x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}}$. We have:

$ (x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}}) \times \hat{\mathbf{i}} = \hat{\mathbf{j}} + \hat{\mathbf{k}} $

Breaking down the cross product:

$x \hat{\mathbf{i}} \times \hat{\mathbf{i}} = 0$

$y \hat{\mathbf{j}} \times \hat{\mathbf{i}} = -y \hat{\mathbf{k}}$

$z \hat{\mathbf{k}} \times \hat{\mathbf{i}} = z \hat{\mathbf{j}}$

Combining, we get:

$ -y \hat{\mathbf{k}} + z \hat{\mathbf{j}} = \hat{\mathbf{j}} + \hat{\mathbf{k}} $

Equating components:

$z = 1$

$-y = 1 \Rightarrow y = -1$

Given $\mathbf{a} \cdot \hat{\mathbf{i}} = 1$:

$ (x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}}) \cdot \hat{\mathbf{i}} = x = 1 $

Thus, we find:

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

Now, the equation of the line passing through the point $\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}$ and parallel to the vector $\mathbf{a}$ is:

$ \mathbf{r} = (\text{Passing through point}) + t (\text{Parallel vector}) $

Substitute:

$ \mathbf{r} = (\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}) + t (\hat{\mathbf{i}} - \hat{\mathbf{j}} + \hat{\mathbf{k}}) $

Simplify:

$ \mathbf{r} = (1 + t) \hat{\mathbf{i}} + (1 - t) \hat{\mathbf{j}} + (1 + t) \hat{\mathbf{k}} $

This is the equation of the line.

Mid-point of $A B$ is $\frac{\mathbf{a}}{\mathbf{2}}+\frac{\mathbf{b}}{2}$ and position vector of $C$ is $\frac{\mathbf{a}}{2}+\frac{\mathbf{b}}{3}$ $\Rightarrow C$ lies inside $\triangle O A B$.

$ \begin{aligned} &|\mathbf{a}-\mathbf{b}|^2+|\mathbf{a}+2 \mathbf{b}|^2=20 \quad \text { [given] } \\ & \begin{aligned} & \Rightarrow|a|^2+|b|^2-2 a \cdot b+|a|^2+4|b|^2 \\ &+4 a \cdot b=20 \\ & \Rightarrow 2|a|^2+5|b|^2+2 a \cdot b=20 \end{aligned} \end{aligned} $

$ \begin{aligned} |\mathbf{a}|=1,|\mathbf{b}| & =2 \\ \mathbf{a} \cdot \mathbf{b} & =|\mathbf{a}||\mathbf{b}| \cos \theta \\ & =2 \cos \theta \\ 2+20+4 \cos \theta & =20 \\ \Rightarrow \quad \cos \theta & =-\frac{1}{2} \\ \Rightarrow \quad \theta & =\frac{2 \pi}{3} \end{aligned} $

$ \begin{aligned} & x=\frac{m_1 x_2+m_2 x_1}{m_1+m_2} \\ & y=\frac{m_1 y_2+m_2 y_1}{m_1+m_2} \end{aligned} $

Let $\mathbf{d}, \mathbf{e}$ and $\mathbf{f}$ are unknown position vectors of $D, E$ and $F$

$ \begin{aligned} \mathbf{d} & =\frac{3 \times \mathbf{c}+1 \times \mathbf{b}}{3+1}=\frac{3 \mathbf{c}+\mathbf{b}}{4}[\text { Divider of } B C] \\ \mathbf{e} & =\frac{4 \times \mathbf{d}+1 \times \mathbf{a}}{4+1} \\ & =\frac{4 \times\left(\frac{3 \mathbf{c}+\mathbf{b}}{4}\right)+\mathbf{a}}{5} \end{aligned} $

$\begin{aligned} & =\frac{3 \mathbf{c}+\mathbf{b}+\mathbf{a}}{5} \quad[\text { divider of } A D] \\ & \mathbf{e}=\frac{3 \times \mathbf{f}+2 \times \mathbf{b}}{3+2} \quad[\text { divider of } B F] \\ \Rightarrow & \frac{3 \mathbf{c}+\mathbf{b}+\mathbf{a}}{5}=\frac{3 \mathbf{f}+2 \mathbf{b}}{5} \\ \Rightarrow & 3 \mathbf{c}+\mathbf{b}+\mathbf{a}=3 \mathbf{f}+2 \mathbf{b} \\ \Rightarrow & 3 \mathbf{c}-\mathbf{b}+\mathbf{a}=3 \mathbf{f} \\ \Rightarrow & \mathbf{f}=\frac{1}{3}(\mathbf{a}-\mathbf{b}+3 \mathbf{c})\end{aligned}$

$ \text { } \begin{aligned} &( \left.\frac{7}{3}+\beta\right) \hat{\mathbf{i}}-\hat{\mathbf{j}}+(\alpha+\gamma) \hat{\mathbf{k}} \\ &= \frac{5}{3}(\alpha \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+\beta(2 \hat{\mathbf{j}}+\hat{\mathbf{k}}) +(\hat{\mathbf{i}}+\gamma \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\ & \Rightarrow(7+3 \beta) \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+(3 \alpha+3 \gamma) \hat{\mathbf{k}} \\ &=5(\alpha \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+\beta(6 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) +(3 \hat{\mathbf{i}}+3 \gamma \hat{\mathbf{j}}+9 \hat{\mathbf{k}}) \\ & \Rightarrow(7+3 \beta) \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+(3 \alpha+3 \gamma) \hat{\mathbf{k}} \\ &=(5 \alpha+3) \hat{\mathbf{i}}+(5+6 \beta+3 \gamma) \hat{\mathbf{j}} +(-5+3 \beta+9) \hat{\mathbf{k}} \end{aligned} $

On comparing corresponding components,

$ \Rightarrow \quad \begin{aligned} 7+3 \beta & =5 \alpha+3 \\ 3 \beta & =5 \alpha-4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ -3 & =5+6 \beta+3 \gamma \end{aligned} $

$ \begin{aligned} \Rightarrow & & 3 \gamma & =-8-6 \beta \\ \Rightarrow & & 3 \gamma & =-10 \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & & 3 \alpha+3 \gamma & =-5+3 \beta+9 \end{aligned} $

$ \Rightarrow \quad 3 \alpha=4+3 \beta-3 \gamma \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

On putting the values from Eqs. (i) and (ii) into Eq. (iii), we get

$ \begin{aligned} & \quad 3 \alpha=4+(5 \alpha-4)-(-10 \alpha) \\ & \Rightarrow \quad 3 \alpha=5 \alpha+10 \alpha \Rightarrow \alpha=0 \\ & \text { From Eq.(i), } 3 \beta=0-4 \\ & \Rightarrow \quad \beta=\frac{-4}{3} \\ & \text { From Eq. (ii), } 3 \gamma=0 \\ & \Rightarrow \quad \gamma=0 \\ & \therefore 5 \alpha-9 \beta+13 \gamma=0-9\left(\frac{-4}{3}\right)+0=12 \end{aligned} $

$ \text { } \begin{aligned} \mathbf{a} & =2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ \mathbf{x} & =\left(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}\right) \mathbf{b} \\ & =\frac{2-1-3}{(1+1+9)} \mathbf{b}=\frac{-2}{11}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\ \mathbf{y} & =\left(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2}\right) \mathbf{a} \\ & =\frac{2-1-3}{4+1+1} \mathbf{a}=-\frac{1}{3}(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ x^2+y^2 & =\frac{4}{121}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \end{aligned} $

$ \begin{aligned} & +\frac{1}{9}(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ = & \frac{4}{121}(1+1+9)+\frac{1}{9}(4+1+1) \\ = & \frac{4}{11}+\frac{2}{3}=\frac{12+22}{33}=\frac{34}{33} \\ \because \cos \theta= & \left|\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\right| \\ \cos \theta= & \left|\frac{2-1-3}{\sqrt{6} \sqrt{11}}\right|=\left|\frac{-2}{\sqrt{6} \sqrt{11}}\right|=\left|\frac{-\sqrt{2}}{\sqrt{3} \sqrt{11}}\right| \\ = & \frac{\sqrt{2}}{\sqrt{33}} \\ \cos ^2 \theta= & \frac{2}{33} \\ \therefore x^2+y^2 & =\frac{34}{33}=17 \times \frac{2}{33}=17 \times \cos ^2 \theta \\ = & 17 \cos ^2 \theta \end{aligned} $

Given, $\mathbf{a}$ and $\mathbf{b}$ determine the base of parallelopiped.

We know,

Volume of parallelopiped $=|[\mathbf{a b c}]|$

$ \begin{array}{lr} \Rightarrow & \text { Area of base × height }=|[\mathbf{a b c}]| \\ \Rightarrow & |\mathbf{a} \times \mathbf{b}| \times \text { height }=|[\mathbf{a b c}]| \\ \Rightarrow & \text { Height }=\frac{|[\mathbf{a b c}]|}{|\mathbf{a} \times \mathbf{b}|} \end{array} $

According to the question,

Using section formula,

$ d=\frac{2 c+3 b}{5}, e=\frac{c+2 a}{3} $

Now, $\hat{\mathbf{p}}=\frac{3\left(\frac{\mathbf{c}+2 \mathbf{a}}{3}\right)+5\left(\frac{2 \mathbf{c}+3 \mathbf{b}}{5}\right)}{8}$

$ \Rightarrow \quad \hat{\mathrm{p}}=\frac{3 \mathrm{c}+2 \mathrm{a}+3 \mathrm{~b}}{8} $

Let $M$ is mid-point of $B C$.

So, position vector of $M=\left(\frac{\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{2}\right)$

Median through vertex $A$ is

$ \begin{aligned} M A & =\mathbf{A}-\mathbf{M} \\ & =(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})-\left(\frac{\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{2}\right) \\ & =\frac{3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-13 \hat{\mathbf{k}}}{2} \end{aligned} $

Vector along the median drawn through vertex $A$ is $\frac{\text { MA }}{\mid \text { MA } \mid}$

$ =\frac{\frac{3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-13 \hat{\mathbf{k}}}{2}}{\frac{\sqrt{9+36+169}}{2}}=\frac{3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-13 \hat{\mathbf{k}}}{\sqrt{214}} $

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

$ \begin{aligned} & |\mathbf{a}-\mathbf{b}|^2+|\mathbf{a}-\mathbf{c}|^2=10 \\ & \Rightarrow|\mathbf{a}|^2+|\mathbf{b}|^2-2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{a}|^2+|\mathbf{c}|^2-2 \mathbf{a c}=10 \\ & \Rightarrow \quad 1+1-2 \mathbf{a} \cdot \mathbf{b}+1+1-2 \mathbf{a} \cdot \mathbf{c}=10 \\ & \Rightarrow \quad \mathbf{a} \cdot \mathbf{b}+\mathbf{a} \cdot \mathbf{c}=-3 \end{aligned} $

$ \begin{aligned} & \text { Statement I }|\mathbf{a}+2 \mathbf{b}|^2+|2 \mathbf{a}+\mathbf{c}|^2 \\ & =|\mathbf{a}|^2+4|\mathbf{b}|^2+4 \mathbf{a} \cdot \mathbf{b}+4|\mathbf{a}|^2+|\mathbf{c}|^2+4 \mathbf{a} \cdot \mathbf{c} \\ & =1+4+4+1+4(\mathbf{a} \cdot \mathbf{b}+\mathbf{a} \cdot \mathbf{c}) \\ & =10+4(-3)=-2 \neq 2 \end{aligned} $

Statement I is false.

$ \begin{aligned} & \text { Statement II }|2 \mathbf{a}+3 \mathbf{b}|^2+|3 \mathbf{a}+2 \mathbf{c}|^2 \\ & \begin{array}{r} =4|\mathbf{a}|^2+9|\mathbf{b}|^2+12 \mathbf{a} \cdot \mathbf{b}+9|\mathbf{a}|^2+4|\mathbf{c}|^2 \\ +12 \mathbf{a} \cdot \mathbf{c} \\ \Rightarrow 4+9+9+4+12(\mathbf{a} \cdot \mathbf{b}+\mathbf{a} \cdot \mathbf{c}) \\ =26+12(-3)=-10 \neq 10 \end{array} \end{aligned} $

Statement II is also false.

Given, $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ are the position vectors of the points $\mathbf{A}$ and $\mathbf{B}$, respectively. Also $\mathbf{C}$ divides $\mathbf{A B}$ in ratio 2: 3 .

∴ Position vector of $\mathbf{C}$ is

$ \begin{aligned} & \frac{2(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})+3(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})}{2+3} \\ & \quad=\frac{2 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}+6 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{5} \\ & \quad=\frac{8}{5} \hat{\mathbf{i}}-\frac{9}{5} \hat{\mathbf{j}}-\frac{7}{5} \hat{\mathbf{k}} \end{aligned} $

As $\mathbf{M}$ is mid-point of $A B$, position vector of $\mathbf{M}$ is

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

5 (position vector of $\mathbf{C}$ ) -2 (position vector of $\mathbf{M}$ )

$ \begin{aligned} & =8 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}-7 \hat{\mathbf{k}}-3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\ & =5 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \end{aligned} $

Let   $\mathbf{A}=\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}$,

        $ \mathbf{B}=-4 \mathbf{a}+5 \mathbf{b}-6 \mathbf{c} $

and   $ \mathbf{C}=x \mathbf{a}-9 \mathbf{b}+z \mathbf{c} $

$\mathbf{A B}=$ Position vector of $\mathbf{B}-$ Position vector of $\mathbf{A}$

$ \begin{aligned} & =-4 a+5 b-6 c-a+2 b-3 c \\ & =-5 a+7 b-9 c \end{aligned} $

$\mathbf{B C}=$ Position vector of $\mathbf{C}$-Position vector of $\mathbf{B}$

$ \begin{aligned} & =x \mathbf{a}-9 \mathbf{b}+z \mathbf{c}+4 \mathbf{a}-5 \mathbf{b}+6 \mathbf{c} \\ & =(x+4) \mathbf{a}-14 \mathbf{b}+(z+6) \mathbf{c} \end{aligned} $

$\mathbf{A}, \mathbf{B}$ and $\mathbf{C}$ are collinears if $\mathbf{A B}=\lambda \mathbf{B C}$

$ \begin{array}{r} -5 a+7 b-9 c=\lambda((x+4) a-14 b \\ +(z+6) c) \end{array} $

$ \begin{aligned} & \therefore \quad 7=-14 \lambda \Rightarrow \lambda=-\frac{1}{2} \\ & \Rightarrow \quad-5=(x+4) \lambda \text { and }-9=\lambda(z+6) \\ & \Rightarrow \quad-5=(x+4)\left(-\frac{1}{2}\right) \text { and } \\ & \quad-9=-\frac{1}{2}(z+6) \\ & \Rightarrow \quad 10=x+4 \text { and } 18=z+6 \\ & \Rightarrow \quad x=6 \text { and } z=12 \\ & \text { Hence, } 2 x-z=2 \times 6-12=0 \end{aligned} $

Given, $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and

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

The component of $\mathbf{b}$ perpendicular to $\mathbf{a}$ is $\mathbf{a} \times \frac{(\mathbf{b} \times \mathbf{a})}{|\mathbf{a}|^2}$.

Now, $\mathbf{b} \times \mathbf{a}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & -3 & 1 \\ \mathbf{i} & -2 & -2\end{array}\right|$

$ \begin{gathered} \hat{\mathbf{i}}(6+2)-\hat{\mathbf{j}}(-4-1)+\hat{\mathbf{k}}(-4+3) \\ =8 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-\hat{\mathbf{k}} \end{gathered} $

$ \begin{aligned} & \therefore \mathbf{a} \times(\mathbf{b} \times \mathbf{a})=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -2 & -2 \\ 8 & 5 & -1 \end{array}\right| \\ & \quad=\hat{\mathbf{i}}(2+10)-\hat{\mathbf{j}}(-1+16)+\hat{\mathbf{k}}(5+16) \\ & \quad=12 \hat{\mathbf{i}}-15 \hat{\mathbf{i}}+21 \hat{\mathbf{k}}=3(4 \hat{\mathbf{i}}-5 \hat{\mathbf{i}}+7 \hat{\mathbf{k}}) \end{aligned} $

$ \begin{aligned} \frac{\mathbf{a} \times(\mathbf{b} \times \mathbf{a})}{|\mathbf{a}|^2} & =\frac{3(4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+7 \hat{\mathbf{k}})}{(1+4+4)} \\ & =\frac{1}{3}(4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}) \end{aligned} $

Given, $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ are the two diagonals of parallelogram. Area of parallelogram $=\frac{1}{2}\left|d_1 \times d_2\right|$

$ \begin{aligned} & =\frac{1}{2}\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 3 & -4 \\ -1 & 2 & 1 \end{array}\right| \\ & =\frac{1}{2}|11 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}| \\ & =\frac{1}{2} \sqrt{121+4+49}=\frac{1}{2} \sqrt{174}=\sqrt{\frac{87}{2}} \end{aligned} $

Let A be taken as origin.

Position vector of $\mathbf{B}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$

Position vector of $\mathbf{C}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$

Position vector of

$ \begin{aligned} & G=\left(\frac{0+2+2}{3}\right) \hat{\mathbf{i}}+\left(\frac{0+2+4}{3}\right) \\ & \hat{\mathbf{j}}+\left(\frac{0+1+4}{3}\right) \hat{\mathbf{k}}=\frac{4}{3} \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\frac{5}{3} \hat{\mathbf{k}} \\ & A G=\frac{4}{3} \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\frac{5}{3} \hat{\mathbf{k}} \\ & |A G|=\sqrt{\frac{16}{9}+4+\frac{25}{9}} \\ & \quad=\sqrt{\frac{16+36+25}{9}}=\sqrt{\frac{77}{9}}=\frac{\sqrt{77}}{3} \end{aligned} $

Now, $\frac{27}{7}|\mathbf{A G}|^2+5$

$ =\frac{27}{7} \times \frac{77}{9}+5=33+5=38 $

$ \begin{aligned} & \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}=\alpha(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) +\beta(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+\gamma(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \end{aligned} $

$ \begin{aligned} \Rightarrow \quad 1 & =\alpha+\beta+2 \gamma \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ -2 & =\alpha+2 \beta-\gamma \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ 5 & =\alpha+3 \beta+\gamma \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

From Eqs. (i) and (ii), we get

$ -\beta+3 \gamma=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

From Eqs. (ii) and (iii), we get

$ \beta+2 \gamma=7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

$ \text { From Eqs. (iv) and (v), we get } $

$ \begin{aligned} \gamma & =2, \text { and } \beta=3 \\ \Rightarrow \quad \alpha & =-6 \\ a^2-\beta^2+\gamma^2 & =36-9+4 \\ & =31 \end{aligned} $

$ \begin{aligned} &\text { Let } \mathbf{a}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \text { and }\\ &\begin{aligned} & \mathbf{b}=a \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+b \hat{\mathbf{k}} \\ & \cos \theta=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} \\ & \frac{2}{3}=\frac{2 a-4+2 b}{\sqrt{4+1+4} \sqrt{a^2+16+b^2}} \\ & \sqrt{a^2+16+b^2}=a+b-2 \quad\\ & a^2+b^2+16=a^2+b^2+4+2 a b -4 a-4 b \\ & 12=2 a b-4 a-4 b \Rightarrow 2 a+2 b+6=a b \\ & 2(a+b+3=a b \end{aligned} \end{aligned} $

Let Position vector of $\mathbf{A}=\hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$, Position vector of $\mathbf{B}$ is $-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$, Position vector of $\mathbf{C}$ is $-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and Position vector of $D$ is $2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+a \hat{\mathbf{k}}$.

$ \begin{aligned} & \mathbf{A B}=-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}} \\ & \mathbf{A C}=-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \mathbf{A D}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+(a+2) \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} & \text { Volume of tetrahedron }=\frac{1}{6}[\mathbf{A B} \mathbf{A C} \mathbf{A D}] \\ & \frac{1}{6}\left|\begin{array}{ccc} -3 & 2 & 0 \\ -2 & -1 & 3 \\ 1 & 3 & a+2 \end{array}\right|=\frac{20}{3} -3(-a-2-9)-2(2-a-4-3) \end{aligned} $

$ \begin{aligned} =-3(-a-11)-2(-2 a-7) & =40 \\ |3 a+33+4 a+14| & =40 \\ |7 a+47| & =40 \\ 7 a+47=40 \text { or } 7 a+47 & =-40 \\ 7 a=-7 \text { or } 7 a & =-87 \end{aligned} $

$ \begin{aligned}⇒ \,\,\,\,\,\,\,\,\,\,\,\,a=-1 \\ or\,\,\,\,\,\,\,\,\,\,\,\,\, a=-\frac{87}{7} \end{aligned} $

Hence, integral value of $a$ is -1 .

$\mathbf{A B}=(7 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})$

$ \begin{aligned} & -(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})=(4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}) \\ C D & =(-7 \hat{\mathbf{i}}-17 \hat{\mathbf{j}}+16 \hat{\mathbf{k}})-(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \\ & =-8 \hat{\mathbf{i}}-14 \hat{\mathbf{j}}+12 \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} \mathbf{A B} \cdot \mathbf{C D} & =-32-98-72=-202 \\ |\mathbf{A B}| & =\sqrt{4^2+7^2+6^2}=\sqrt{101} \\ |\mathbf{C D}| & =\sqrt{8^2+(14)^2+(12)^2}=\sqrt{404} \end{aligned} $

Let $\theta$ be the required angle.

$ \begin{aligned} \therefore \cos \theta & =\frac{\mathbf{A B} \cdot \mathbf{C D}}{|\mathbf{A B}||\mathbf{C D}|}=\frac{-202}{\sqrt{101} \sqrt{404}} \\ & =\frac{-202}{2 \times 101} \\ \cos \theta & =-1 ; \quad \therefore \quad \theta=\pi \end{aligned} $

$ \begin{aligned} \mathbf{A C} & =-6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \\ \mathbf{A D} & =-3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\ \because \quad \mathbf{A B} & =x \mathbf{A C}+y \mathbf{A D} \end{aligned} $

$\therefore \mathbf{A B}, \mathbf{A C}$ and $\mathbf{A D}$ are coplanar.

$ \begin{aligned} & \Rightarrow \quad\left|\begin{array}{ccc} \lambda-2 & 4 & 5 \\ -6 & 2 & 3 \\ -3 & -3 & 4 \end{array}\right|=0 \\ & \Rightarrow 17(\lambda-2)-4(-15)+5(24)=0 \\ & \Rightarrow \quad \lambda=2-\frac{180}{17} \\ & \therefore \quad 17(\lambda+9)=17\left(2-\frac{180}{17}+9\right)=7 \end{aligned} $

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

$ \begin{gathered} \mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\ \mathbf{a} \cdot \mathbf{b}=x+y+z=1 \end{gathered} $

and $\cos (\mathbf{a} \cdot \mathbf{b})=\frac{1}{3}$

$ \begin{aligned} & \text { Also, } \mathbf{a} \cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}| \cdot \frac{1}{3}=1 \\ & \Rightarrow \sqrt{3} \cdot|\mathbf{b}| \cdot \frac{1}{3}=1 \Rightarrow|\mathbf{b}|=\sqrt{3} \\ & \therefore x^2+y^2+z^2=3 \end{aligned} $

Possible values of $(x, y, z)$ are

$ (1,1,-1),(1,-1,1) \text { and }(-1,1,1) $

Hence, three vectors are possible.

$|a|=\sqrt{2^2+2^2+p^2}=\sqrt{8+p^2}$ and $|b|=7$

$\mathbf{a} \cdot \mathbf{b}=4 \,\,\,\,[given]$

$ \begin{aligned} & \Rightarrow \quad|\mathbf{a}||\mathbf{b}| \cos \theta=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { Again, } \quad|\mathbf{a} \times \mathbf{b}|=5 \sqrt{17} \\ & \Rightarrow \quad|\mathbf{a}||\mathbf{b}| \sin \theta=5 \sqrt{17} \\ & \Rightarrow \quad \frac{4}{\cos \theta} \times \sin \theta=5 \sqrt{17} \quad \text { [From Eq. (i)] } \\ & \Rightarrow \quad \tan \theta=\frac{5 \sqrt{17}}{4} \Rightarrow \cos \theta=\frac{4}{21} \end{aligned} $

$ \begin{aligned} & & \frac{4}{|\mathbf{a}||\mathbf{b}|} & =\frac{4}{21}\,\,\,\, \text { [From Eq. (i)] }\\ \Rightarrow & & |\mathbf{a}||\mathbf{b}| & =21 \\ \Rightarrow & & \sqrt{8+p^2} \times 7 & =21 \Rightarrow 8+p^2=9 \end{aligned} $

$ \begin{aligned} \Rightarrow & p^2 =1 \\ \therefore & p = \pm 1 \end{aligned} $

Let $A, B, C, D, E, P$ have position vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}, \mathbf{e}$ and $\mathbf{p}$ respectively.

Given, $D$ and $E$ divide the line segment $B C$ and $A E$ in the ratio $2: 1$.

By section formula,

$ \begin{aligned} & \mathbf{d}=\frac{2 c+b}{2+1} \Rightarrow 3 d=2 c+b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & e=\frac{c+2 a}{2+1} \Rightarrow 3 e=c+2 a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eq. (i), $3 \mathbf{d}-\mathbf{b}=\mathbf{2} \mathbf{c}$

From Eq. (ii), $3 \mathbf{e}-2 \mathbf{a}=\mathbf{c}$

$\therefore \quad 6 e-4 a=2 c$

Equating both values of 2 c ,

$ \begin{aligned} & & 3 \mathrm{~d}-\mathrm{b} & =6 \mathrm{e}-4 \mathrm{a} \\ \Rightarrow & & 3 \mathrm{~d}+4 \mathrm{a} & =6 \mathrm{e}+\mathrm{b} \\ \Rightarrow & & \frac{3 \mathrm{~d}+4 \mathrm{a}}{3+4} & =\frac{6 \mathrm{e}+\mathrm{b}}{6+1} \end{aligned} $

LHS is the position vector of the point which divides segment $A D$ internally in the ratio $3: 4$. RHS is the position

vector of the point which divides segment $B E$ internally in the ratio $6: 1$. But $P$ is the point of intersection of $A D$ and $B E$.

$\therefore P$ divides $A D$ internally in the ratio 3 : 4 and $P$ divides $B E$ internally in the ratio $6: 1$.

Let position vector of

$ \begin{aligned} & A: \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & B: 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\ & C:-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-5 \hat{\mathbf{k}} \\ & D: a \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\ & A B: \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-7 \hat{\mathbf{k}} \\ & A C:-4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-8 \hat{\mathbf{k}} \\ & A D:(a-1) \hat{\mathbf{i}}+0 \hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} $

Given $\mathbf{A}, \mathbf{B}, \mathbf{C}$ and $\mathbf{D}$ are coplanar.

$\left[\begin{array}{lll}\mathbf{A B} & \mathbf{A C} & \mathbf{A D}\end{array}\right]=0$

$ \left|\begin{array}{ccc} 1 & 5 & -7 \\ -4 & 3 & -8 \\ a-1 & 0 & 1 \end{array}\right|=0 $

$ \begin{aligned} & \Rightarrow(a-1)(-40+21)+0+1(3+20)=0 \\ & \Rightarrow \quad-19(a-1)+23=0 \\ & \Rightarrow \quad-19 a+19+23=0 \Rightarrow a=\frac{42}{19} \end{aligned} $

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

$\mathbf{a}$ be a vector in the plane containing vectors

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

$\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=0$

$ \begin{aligned} &\left|\begin{array}{ccc} a & b & c \\ 1 & 2 & 1 \\ 2 & -1 & 1 \end{array}\right|=0 \\ & \Rightarrow 3 a+b-5 c=0 \quad \ldots(\mathbf{i}) \\ & \Rightarrow \mathbf{a} \perp(\mathbf{i}+\mathbf{j}+3 \mathbf{k}) \Rightarrow \mathbf{a} \cdot(\mathbf{i}+\mathbf{j}+3 \mathbf{k})=0 \end{aligned} $

$ \begin{aligned} &a+b+3 c=0\\ &\text { Projection of } \mathbf{a} \text { on } \mathbf{b} \text { is } 3 \sqrt{6} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\text {. } \end{aligned} $

$ \begin{aligned} & \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}=3 \sqrt{6} \Rightarrow \frac{a+2 b+c}{\sqrt{1+4+1}}=3 \sqrt{6} \\ & a+2 b+c=18 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

on subtracting Eq. (ii) from Eq. (i),

$ 2 a-8 c=0 \Rightarrow a=4 c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

on multiplying 2 in Eq. (ii) and subtracting

Eq. (iii), we get

$ a+5 c=-18 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

From Eqs. (iv) and (v), $c=-2$ and $a=-8$

On putting in Eq. (iii), $b=14$

$ \begin{gathered} \mathbf{a}=-8 \hat{\mathbf{i}}+14 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ |a|^2=8^2+14^2+2^2=64+196+4=264 \end{gathered} $

$ \begin{aligned} & \text { Let } \mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \mathbf{b}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \text { c }=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & \mathbf{d}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & l=\mathbf{b} \cdot \mathbf{c}=1-6-2=-7 \\ & m=\mathbf{c} \cdot \mathbf{a}=1+3-2=2 \\ & {[m \mathbf{b}+l \mathbf{a} \mathbf{b} \mathbf{d}]=[m \mathbf{b} \mathbf{b} \mathbf{d}]+[l \mathbf{l} \mathbf{a} \mathbf{b} \mathbf{d}]} \\ & \Rightarrow[\mathrm{labd}]=l[\mathrm{abd}] \\ & =-7\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -2 & 1 \\ 2 & 1 & -1 \end{array}\right|=-7 \\ & {[1+3+5]=-63} \end{aligned} $

$ \text { ∴ Vector normal to plane } \mathbf{n}=\mathbf{b} \times \mathbf{c} $

$ \begin{aligned} & \therefore \quad \mathbf{n}=\frac{\mathbf{b} \times \mathbf{c}}{|\mathbf{b} \times \mathbf{c}|} \\ & \text { Now } \mathbf{A r} \cdot \mathbf{n}=0 \\ & (\mathbf{r}-\mathbf{a}) \cdot \frac{(\mathbf{b} \times \mathbf{c})}{|\mathbf{b} \times \mathbf{c}|}=0 \\ & \Rightarrow \quad \mathbf{r} \cdot \frac{\mathbf{b} \times \mathbf{c}}{|\mathbf{b} \times \mathbf{c}|}=\mathbf{a} \cdot \frac{(\mathbf{b} \times \mathbf{c})}{|\mathbf{b} \times \mathbf{c}|} \\ & \quad \mathbf{r} \cdot \frac{(\mathbf{b} \times \mathbf{c})}{|\mathbf{b} \times \mathbf{c}|}=\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{|\mathbf{b} \times \mathbf{c}|} \end{aligned} $