Vector Algebra

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

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

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

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

Since $\alpha=\sqrt{3}$

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

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

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

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

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

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

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

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

Let the normal vector of plane $\pi_1$ be

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

Let the normal vector of plane $\pi_2$ be

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

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

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

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

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

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

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

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

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

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

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

Shortest distance between two lines

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

⇒ lines intersect

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

∴ Centroid of tetrahedron $A B C D$ is

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

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

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

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

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

From by Eq. (iii), we get

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

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

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

Vector $\mathbf{d}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$

$ \begin{aligned} \therefore \quad \mathbf{d} & =\lambda(\mathbf{a} \times \mathbf{b}) \\ \mathbf{a} \times \mathbf{b} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 1 \\ 1 & -2 & -2 \end{array}\right|=4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} $

Let angle between $(\mathbf{a} \times \mathbf{b})$ and $\mathbf{c}$ be $\theta$.

$ \begin{aligned} \therefore \cos \theta & =\frac{(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}}{|\mathbf{a} \times \mathbf{b}||\mathbf{c}|} \\ & =\frac{(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \cdot(6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})}{\sqrt{26} \times 7} \\ & =\frac{24+9+2}{\sqrt{26} \times 7}=\frac{5}{\sqrt{26}} \end{aligned} $

$ \begin{aligned} & \therefore \sin \theta=\frac{1}{\sqrt{26}} \\ & \text { Now, }|\mathbf{d} \times \mathbf{c}|=14 \\ & |\mathbf{d} \| \mathbf{c}| \sin \theta=14 \\ & \lambda|\mathbf{a} \times \mathbf{b} \| \mathbf{c}| \sin \theta=14 \quad[\because \hat{\mathbf{d}}=\lambda(\mathbf{a} \times \mathbf{b})] \\ & \lambda \times \sqrt{26} \times 7 \times \frac{1}{\sqrt{26}}=14 \end{aligned} $

$ \begin{aligned} & \lambda=2 \\ & \begin{aligned} \therefore|\mathbf{d} \cdot \mathbf{c}| & =|\lambda(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}| \\ & =2(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \cdot(6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ & =2(24+9+2)=70 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \mathbf{a}=(x+2 y-3) \hat{\mathbf{i}}+(2 x-y+3) \hat{\mathbf{j}} \\ & \quad \mathbf{b}=(3 x-2 y) \hat{\mathbf{i}}+(x-y+1) \hat{\mathbf{j}} \\ & \because \mathbf{a}=2 \mathbf{b} \\ & \therefore x+2 y-3=2(3 x-2 y) \\ & \quad x+2 y-3=6 x-4 y \\ & \quad 5 x-6 y+3=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} \text { And } & 2 x-y+3=2(x-y+1) \\ & 2 x-y+3=2 x-2 y+2 \\ & y=-1 \,\,\,\,\,\,\,(Using\,\,\,\, Eq. (i))\\ \therefore \quad & x=\frac{-9}{5} \\ \therefore \quad & y-5 x \\ & =-1+9=8 \end{array}\\ &\text { } \end{aligned} $

Position vectors of $A, B, C$ and $D$ are

$ \begin{aligned} & \mathbf{a}=7 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}} \\ & \mathbf{b}=\hat{\mathbf{i}}-6 \hat{\mathbf{j}}+10 \hat{\mathbf{k}} \\ & \mathbf{c}=-\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\ & \mathbf{d}=5 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} \text { respectively. } \end{aligned} $

∵ Let point of intersection of $\mathbf{A C}$ and $\mathbf{B D}$ be $P$

$ \begin{aligned} &\text { ∴ Position of vector of } P \text { is }\left(\frac{\lambda \mathbf{c}+\mathbf{a}}{\lambda+1}\right) \text { or }\\ &\begin{aligned} & \frac{\mu \mathbf{d}+\mathbf{b}}{\mu+1} \\ & \therefore \frac{-\lambda \hat{\mathbf{i}}-3 \lambda \hat{\mathbf{j}}+4 \lambda \hat{\mathbf{k}}+7 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}}{\lambda+1} \\ & =\frac{5 \mu \hat{\mathbf{i}}-\mu \hat{\mathbf{j}}+\mu \hat{\mathbf{k}}+\hat{\mathbf{i}}-6 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}}{\mu+1} \\ & \frac{-\lambda+7}{\lambda+1}=\frac{5 \mu+1}{\mu+1} \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow-\lambda \mu+7 \mu-\lambda+7=5 \lambda \mu+\lambda+5 \mu+1 \\ & 6 \lambda \mu+2 \lambda-2 \mu=6 \\ & 3 \lambda \mu+\lambda-\mu=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \frac{-3 \lambda-4}{\lambda+1}=\frac{-\mu-6}{\mu+1} \\ & \Rightarrow-3 \lambda \mu-4 \mu-3 \lambda-4=-\lambda \mu-6 \lambda-\mu-6 \\ & 2 \lambda \mu-3 \lambda+3 \mu=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \frac{4 \lambda+7}{\lambda+1}=\frac{\mu+10}{\mu+1} \\ & \Rightarrow 4 \lambda \mu+7+7 \mu+4 \lambda=\lambda \mu+10 \lambda+\mu+10 \\ & 3 \lambda \mu-6 \lambda+6 \mu=3 \\ & \lambda \mu-2 \lambda+2 \mu=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & \lambda(\mu-2)=1-2 \mu \\ & \lambda=\frac{1-2 \mu}{\mu-2} . \end{aligned} $

By Eq. (i)

$ \begin{aligned} & 3 \mu\left(\frac{1-2 \mu}{\mu-2}\right)+\frac{1-2 \mu}{\mu-2}-\mu=3 \\ & \Rightarrow 3 \mu-6 \mu^2+1-2 \mu-\mu^2+2 \mu=3 \mu-6 \\ & \Rightarrow \mu= \pm 1 \\ & \text { If } \mu=1, \quad \lambda=1 \text { and } \mu=-1, \quad \lambda=-1 \end{aligned} $

∴ Position vector of

$ \begin{aligned} & P=\frac{(-\lambda+7) \hat{\mathbf{i}}+(-3 \lambda-4) \hat{\mathbf{j}}+(4 \lambda+7) \hat{\mathbf{k}}}{\lambda+1} \\ & \begin{aligned} \therefore p+q+r & =\frac{-\lambda+7-3 \lambda-4+4 \lambda+7}{\lambda+1} \\ & =\frac{10}{\lambda+1}=5(\text { At } \lambda=1) \end{aligned} \end{aligned} $

Given, $\mathbf{a}=\hat{\mathbf{i}}+\sqrt{11} \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$,

$ \mathbf{b}=\hat{\mathbf{i}}+\sqrt{11 \hat{\mathbf{j}}}-10 \hat{\mathbf{k}} $

∴ component of $\mathbf{b}$ perpendicular to $\mathbf{a}$ is

$ \begin{aligned} & \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \hat{\mathbf{a}} \\ & =(\hat{\mathbf{i}}+\sqrt{11 \hat{\mathbf{j}}}-10 \hat{\mathbf{k}}) \end{aligned} $

$ -\left[\frac{(\hat{\mathbf{i}}+\sqrt{11 \hat{\mathbf{j}}}-2 \hat{\mathbf{k}})}{\sqrt{16}} \cdot(\hat{\mathbf{i}}+\sqrt{11 \hat{\mathbf{j}}}-10 \hat{\mathbf{k}})\right] \frac{\hat{\mathbf{i}}+\sqrt{11} \hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{\sqrt{16}}$

$ \begin{aligned} & =\hat{\mathbf{i}}+\sqrt{11} \hat{\mathbf{j}}-10 \hat{\mathbf{k}}-\frac{(1+11+20)}{4}.\frac{(\hat{\mathbf{i}}+\sqrt{11 \hat{\mathbf{j}}}-2 \hat{\mathbf{k}})}{4} \\ & =\hat{\mathbf{i}}+\sqrt{11 \hat{\mathbf{j}}+\sqrt{11} \hat{\mathbf{j}}-2 \hat{\mathbf{k}})} \\ & =-\hat{\mathbf{k}}-2(\hat{\mathbf{i}}+\sqrt{11 \hat{\mathbf{k}}}-2 \hat{\mathbf{k}}) \\ & =-(\hat{\mathbf{i}}+\sqrt{11} \hat{\mathbf{j}}-6 \hat{\mathbf{k}}+6 \hat{\mathbf{k}}) \end{aligned} $

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

$ \begin{aligned} & \because \mathbf{a} \cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}| \cos \theta \\ & (2-2+2 p)=\sqrt{9} \sqrt{5+p^2} \cdot \cos 60^{\circ} \\ & \quad \quad 2 p=3 \sqrt{5+p^2} \frac{1}{2} \\ & \Rightarrow \quad 4 p=3 \sqrt{5+p^2} \\ & \Rightarrow \quad 16 p^2=9\left(5+p^2\right) \\ & \Rightarrow \quad 16 p^2=45+9 p^2 \\ & \Rightarrow \quad 7 p^2=45 \Rightarrow p^2=\frac{45}{7} \\ & \Rightarrow \quad p=\frac{3 \sqrt{5}}{\sqrt{7}} \end{aligned} $

$ \begin{aligned} &\text { Let position vector of } A, B, C \text { and } D \text { be respectively } \mathbf{a}, \mathbf{b}, \mathbf{c} \text { and } \mathbf{d}\\ &\begin{aligned} & \therefore \quad A B=b-a, \quad C B=b-c \\ & \quad C D=d-c, \quad A D=d-a \\ & \therefore A B+C B+C D+A D \\ & \quad=2 b+2 d-2 a-2 c \\ & \quad=2(b+d-a-c)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} \end{aligned} $

And let position vector of $E$ and $F$ are $\mathbf{e}$ and $\mathbf{f}$

$ \begin{array}{ll} \therefore & \mathbf{e}=\frac{\mathbf{a}+\mathbf{c}}{2} \\ \Rightarrow & \mathbf{a}+\mathbf{c}=2 \mathbf{e} \\ \text { And } & \mathbf{f}=\frac{\mathbf{b}+\mathbf{d}}{2} \Rightarrow \mathbf{b}+\mathbf{d}=2 \mathbf{f} \end{array} $

$\therefore$ From Eq. (i), we get

$ \begin{aligned} & \mathbf{A B}+\mathbf{C B}+\mathbf{C D}+\mathbf{A D}=2(2 \mathbf{f}-2 \mathbf{c}) \\ & =4(\mathbf{f}-\mathbf{e})=4 \mathbf{E F} \end{aligned} $

We have, $\mathbf{O A}=2 \mathbf{a}+3 \mathbf{b}-\mathbf{c}$

$ \quad \begin{aligned} & O B=a-2 b+3 c \\ & O C=3 a+4 b-2 c \\ & O D=a-6 b+6 c \\ \therefore\,\, & A B=-a-5 b+4 c \\ & A C=a+b-c \\ & A D=-a-9 b+7 c \end{aligned} $

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

$ \begin{aligned} & =\left|\begin{array}{rrr} -1 & -5 & 4 \\ 1 & 1 & -1 \\ -1 & -9 & 7 \end{array}\right| \\ & =(-1)(7-9)+5(7-1)+4(-9+1) \\ & =2+30-32=0 \end{aligned} $

Given, point are coplanar.

$ \begin{aligned} &\text { Given, } a=|\mathbf{a}|, b=|\mathbf{b}|\\ &\begin{aligned} & \left(\frac{\mathbf{a}}{a^2}-\frac{\mathbf{b}}{b^2}\right)^2=\left(\frac{\mathbf{a}}{a^2}-\frac{\mathbf{b}}{b^2}\right)\left(\frac{\mathbf{a}}{a^2}-\frac{\mathbf{b}}{b^2}\right) \\ & =\left|\frac{\mathbf{a}}{a^2}\right|^2+\left|\frac{\mathbf{b}}{b^2}\right|^2-2\left(\frac{\mathbf{a}}{a^2}\right) \cdot\left(\frac{\mathbf{b}}{b^2}\right) \\ & =\frac{|\mathbf{a}|^2}{a^4}+\frac{|\mathbf{b}|^2}{b^4}-\frac{2(\mathbf{a} \cdot \mathbf{b})}{a^2 b^2} \\ & =\frac{1}{a^2}+\frac{1}{b^2}-\frac{2(\mathbf{a} \cdot \mathbf{b})}{a^2 b^2}=\frac{a^2+b^2-2(\mathbf{a} \cdot \mathbf{b})}{a^2 b^2} \\ & =\frac{(\mathbf{a}-\mathbf{b})^2}{a^2 b^2}=\left(\frac{\mathbf{a}-\mathbf{b}}{a b}\right)^2 \end{aligned} \end{aligned} $

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

$ \begin{aligned} \mathbf{a} \cdot \mathbf{b} & =|\mathbf{a}||\mathbf{b}| \cos \frac{\pi}{3} \\ & =\text { (l) (l) }\left(\frac{1}{2}\right)=\frac{1}{2} \\ \mathbf{b} \cdot \mathbf{c} & =|\mathbf{b}||\mathbf{c}| \cos \frac{\pi}{3} \\ & =\text { (l) (l) ( } \frac{1}{2} \text { ) }=\frac{1}{2} \end{aligned} $

And $\mathbf{c} \cdot \mathbf{a}=|\mathbf{c}||\mathbf{a}| \cos \frac{\pi}{3}=(1)(1)\left(\frac{1}{2}\right)=\frac{1}{2}$

$ \begin{aligned} {[\mathbf{a} \mathbf{b} \mathbf{c}] } & =\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c}) \\ & =|\mathbf{a}||\mathbf{b} \times \mathbf{c}| \cos \frac{\pi}{6} \end{aligned} $

$ =\text { (l) }|\mathbf{b} \times \mathbf{c}|\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{2}|\mathbf{b} \times \mathbf{c}| $

Now, $|\mathbf{b} \times \mathbf{c}|=|\mathbf{b}||\mathbf{c}| \sin \frac{\pi}{3}$

$ \begin{aligned} & =(1)(1)\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{2} \\ \therefore \quad[\mathbf{a b c}] & =\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}=\frac{3}{4} \end{aligned} $

Given, $x \mathbf{a}+y \mathbf{b}+z \mathbf{c}=p(\mathbf{b} \times \mathbf{c})+q(\mathbf{c} \times \mathbf{a}) +\mathbf{r}(\mathbf{a} \times \mathbf{b})$

$ \begin{array}{ll} \therefore & x(\mathbf{a} \cdot \mathbf{a})+y(\mathbf{b} \cdot \mathbf{a})+z(\mathbf{c} \cdot \mathbf{a}) \\ & =p((\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a}) \\ \Rightarrow & x+y\left(\frac{1}{2}\right)+z\left(\frac{1}{2}\right)=p[\mathbf{a} \mathbf{b} \mathbf{c}] \\ \Rightarrow & 2 x+y+z=\frac{3}{2} p \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Similarly,

$ \begin{aligned} & x(\mathbf{a} \cdot \mathbf{b})+y(\mathbf{b} \cdot \mathbf{b})+z(\mathbf{c} \cdot \mathbf{b})=q((\mathbf{c} \times \mathbf{a}) \cdot \mathbf{b}) \\ & \Rightarrow \quad x\left(\frac{1}{2}\right)+y+z\left(\frac{1}{2}\right)=q[\mathbf{a} \mathbf{b} \mathbf{c}] \\ & \Rightarrow \quad x+2 y+z=\frac{3}{2} q \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

And $x(\mathbf{a} \cdot \mathbf{c})+y(\mathbf{b} \cdot \mathbf{c})+z(\mathbf{c} \cdot \mathbf{c})=r[\mathbf{a} \mathbf{b} \mathbf{c}]$

$ \begin{aligned} & \Rightarrow \quad x\left(\frac{1}{2}\right)+y\left(\frac{1}{2}\right)+z=r\left(\frac{3}{4}\right) \\ & \Rightarrow \quad x+y+2 z=\frac{3}{2} r \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

On adding Eqs. (i), (ii) and (iii), we get

$ \begin{aligned} & 4(x+y+z)=\frac{3}{2}(p+q+r) \\ \Rightarrow \quad & \frac{x+y+z}{p+q+r}=\frac{3}{8} \end{aligned} $

We have, vertices of a tetrahedron $O(0,0,0), A(3,1,4), B(1,3,2)$ and $C(0,4,-2)$

$\therefore$ Centroid of tetrahedron, $G$

$ \begin{aligned} & =\left(\frac{0+3+1+1+0}{4}, \frac{0+1+3+4}{4},\right. \left.\frac{0+4+2-2}{4}\right) \\ & =\left(\frac{4}{4}, \frac{8}{4}, \frac{4}{4}\right)=(1,2,1) \end{aligned} $

and vertices of face $A B C$

$ A(3,1,4), B(1,3,2) \text { and } C(0,4,-2) $

$\therefore$ Centroid of face $A B C$,

$ \begin{aligned} G_1 & =\left(\frac{3+1+0}{3}, \frac{1+3+4}{3}, \frac{4+2-2}{3}\right) \\ & =\left(\frac{4}{3}, \frac{8}{3}, \frac{4}{3}\right) \end{aligned} $

Let $P(x, y)$ be the point which divides $G G_1$ in the ratio $1: 2$ is

$ \left(\frac{1 \times \frac{4}{3}+2 \times 1}{1+2}, \frac{1 \times \frac{8}{3}+2 \times 2}{1+2}, \frac{1 \times \frac{4}{3}+2 \times 1}{1+2}\right) $

$ \begin{aligned} & \therefore P(x, y, z)=\left(\frac{4+6}{3(3)}, \frac{8+12}{3(3)}, \frac{4+6}{3(3)}\right) \\ & \Rightarrow P(x, y, z)=\left(\frac{10}{9}, \frac{20}{9}, \frac{10}{9}\right) \end{aligned} $

Given, $\mathbf{A}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{B}=7 \hat{\mathbf{i}}-\hat{\mathbf{k}}$,

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

If $P$ divides $A B$ in the ratio $m: n$, then its harmonic conjugate $Q$ divides $A B$ in the ratio $-m: n$.

Now, $\mathbf{A B}=\mathbf{B}-\mathbf{A}$

$ \begin{aligned} & =(7-1) \hat{\mathbf{i}}+(0-2) \hat{\mathbf{j}}+(-1-3) \hat{\mathbf{k}} \\ & =6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \end{aligned} $

Let $P$ divides $A B$ in the ratio $m: n$, so

$ \mathbf{P}=\frac{n \mathbf{A}+m \mathbf{B}}{m+n} $

$ \begin{array}{ll} \Rightarrow \quad & -2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}} \\ & =\frac{n(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+m(7 \hat{\mathbf{i}}-\hat{\mathbf{k}})}{m+n} \\ \Rightarrow \quad & (m+n)(-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \\ & =n(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+m(7 \hat{\mathbf{i}}-\hat{\mathbf{k}}) \end{array} $

Comparing the coefficients of $\hat{\mathbf{i}}, \hat{\mathbf{j}}$ and $\hat{\mathbf{k}}$ on both sides

$ \begin{aligned} \Rightarrow & & -2(m+n) & =n(1)+m(7) \\ \Rightarrow & & -2 m-2 n & =n+7 m \\ \Rightarrow & & -9 m & =3 n \\ \Rightarrow & & n & =-3 m \\ \Rightarrow & & \frac{m}{n} & =\frac{1}{-3} \\ \therefore & & \frac{-m}{n} & =\frac{-1}{-3}=\frac{1}{3} \end{aligned} $

Now using section formula,

$ \begin{aligned} Q & =\frac{3 A+1 B}{3+1} \\ & =\frac{3(\hat{i}+2 \hat{j}+3 \hat{k})+(7 \hat{i}-\hat{k})}{4} \\ & =\frac{10 \hat{i}+6 \hat{j}+8 \hat{k}}{4}=\frac{5}{2} \hat{i}+\frac{3}{2} \hat{j}+2 \hat{k} \end{aligned} $

So, sum of scalar components of

$ \begin{aligned} Q & =\frac{5}{2}+\frac{3}{2}+2 \\ & =\frac{8}{2}+2=4+2=6 \end{aligned} $

Let the points be $\mathbf{A}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{B}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and plane through points are $\mathbf{P}_1=\hat{\mathbf{i}}, \mathbf{P}_2=2 \hat{\mathbf{j}}, \mathbf{P}_3=3 \hat{\mathbf{k}}$

Let the point on the line be

$ \begin{aligned} & \mathbf{r}(t)=\mathbf{A}+t(\mathbf{B}-\mathbf{A}) \\ & \Rightarrow \mathbf{r}(t)=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+t[(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & \quad \quad(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})] \\ & \quad=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+t(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ & \quad=(1+t) \hat{\mathbf{i}}+(2-3) \hat{\mathbf{j}}+(1-2 t) \hat{\mathbf{k}} \end{aligned} $

Now, vectors in the plane are

$ \begin{aligned} & \mathbf{v}_1=\mathbf{P}_2-\mathbf{P}_1=2 \hat{\mathbf{j}}-\hat{\mathbf{i}} \\ & \mathbf{v}_2=\mathbf{P}_3-\mathbf{P}_1=3 \hat{\mathbf{k}}-\hat{\mathbf{i}} \end{aligned} $

Normal vectors, $\mathbf{n}=\mathbf{v}_1 \times \mathbf{v}_2$

$ \begin{aligned} & \begin{array}{r} \Rightarrow\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -1 & 2 & 0 \\ -1 & 0 & 3 \end{array}\right|=(6-0) \hat{\mathbf{i}}-(-3-0) \hat{\mathbf{j}} +(0+2 \hat{\mathbf{k}} \end{array} \\ & \Rightarrow \quad 6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned} $

Using point $\mathbf{P}_1=\hat{\mathbf{i}}$ to get plane equation is

$ \mathbf{n} \cdot(\mathbf{r}-\hat{\mathbf{i}})=0 $

$ \begin{array}{ll} \Rightarrow & \langle 6,3,2\rangle \cdot\langle x-1, y, z\rangle=0 \\ \Rightarrow & 6 x-6+3 y+2 z=0 \\ \Rightarrow & 6 x+3 y+2 z=6 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Put the value of $\mathbf{r}(t)=(1+t, 2-3 t, 1-2 t)$ into the plane Eq. (i), we get

$ \begin{aligned} & 6 x+3 y+2 z=6 \\ & \Rightarrow 6(1+t)+3(2-3 t)+2(1-2 t)=6 \\ & \Rightarrow 6+6 t+6-9 t+2-4 t=6 \\ & \Rightarrow-7 t+14=6 \Rightarrow 7 t=8 \\ & \Rightarrow \quad t=\frac{8}{7} \\ & \text { So, } \mathbf{r}\left(\frac{8}{7}\right)=\left(1+\frac{8}{7}\right) \hat{\mathbf{i}}+\left(2-3 \times \frac{8}{7}\right) \hat{\mathbf{j}} +\left(1-2 \times \frac{8}{7}\right) \hat{\mathbf{k}} \\ & \Rightarrow \frac{15}{7} \hat{\mathbf{i}}-\frac{10}{7} \hat{\mathbf{j}}-\frac{9}{7} \hat{\mathbf{k}} \end{aligned} $

Given, $|\mathbf{a}|=5,|\mathbf{b}|=12|\mathbf{a}-\mathbf{b}|=13$

Now, $|\mathbf{a}-\mathbf{b}|^2=13^2=169$

$ \begin{aligned} & \Rightarrow \quad|\mathbf{a}|^2+|\mathbf{b}|^2-2 \mathbf{a} \cdot \mathbf{b}=169 \\ & \Rightarrow \quad 5^2+12^2-2 \mathbf{a} \cdot \mathbf{b}=169 \\ & \Rightarrow \quad-2 \mathbf{a} \cdot \mathbf{b}=169-25-144=0 \\ & \Rightarrow \quad \mathbf{a} \cdot \mathbf{b}=0 \end{aligned} $

So, $\mathbf{a} \perp \mathbf{b}$

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

$ \begin{array}{ll} \Rightarrow \quad & 4 \mathbf{a} \cdot \mathbf{a}+4 \mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{b} \\ & =4|\mathbf{a}|^2+4 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^2 \\ & =4(5)^2+4 \times 0+(1)^2 \\ & =100+0+144=244 \\ \therefore \quad & |\mathbf{2} \mathbf{a}+\mathbf{b}|=\sqrt{244}=2 \sqrt{61} \end{array} $

$ \begin{aligned} &\text {Given, } \mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \text { and }\\ &\begin{aligned} & \mathbf{b}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \mathbf{a}+2 \mathbf{b}=(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+2(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ & \Rightarrow \quad(\mathrm{l}+4) \hat{\mathbf{i}}+(-2+2) \hat{\mathbf{j}}+(-2+4) \hat{\mathbf{k}} \\ & \Rightarrow \quad 5 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \quad 3 \mathbf{a}-\mathbf{b}=3(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})-(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ & \Rightarrow \quad(3-2) \hat{\mathbf{i}}+(-6-1) \hat{\mathbf{j}}+(-6-2) \hat{\mathbf{k}} \\ & \Rightarrow \quad \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-8 \hat{\mathbf{k}} \\ & \text { Now, }(\mathbf{a}+2 \mathbf{b}) \times(3 \mathbf{a}-\mathbf{b})=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 5 & 0 & 2 \\ 1 & -7 & -8 \end{array}\right| \\ & \Rightarrow(0-(-14)) \hat{\mathbf{i}}-(-40-2) \hat{\mathbf{j}}+(-35-0) \hat{\mathbf{k}} \\ & \Rightarrow 14 \hat{\mathbf{i}}+42 \hat{\mathbf{j}}-35 \hat{\mathbf{k}} \end{aligned} \end{aligned} $

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} $