Vector Algebra

$ \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} $

$ C=\left(\frac{344}{5}, \frac{6-6}{5}, \frac{-9+2}{5}\right) $

$ \begin{array}{l} C\left(\frac{7}{5}, 0,-\frac{7}{5}\right) \\\\ O C=\frac{7}{5} \hat{\mathbf{i}}-\frac{7}{5} \hat{\mathbf{k}} \\\\ O D=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\\\ \mathbf{C D}=\frac{8}{5} \hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{17}{5} \hat{\mathbf{k}} \\\\ \begin{aligned} \mathbf{C D} & =\sqrt{\frac{64}{25}+1+\frac{289}{25}}=\sqrt{\frac{378}{25}}=\frac{3 \sqrt{42}}{5} \\\\ \mathbf{C D} & =\frac{\mathbf{C D}}{|\mathbf{C D}|}=\frac{\frac{1}{5}(8 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+17 \hat{\mathbf{k}})}{\frac{3 \sqrt{42}}{5}} \\\\ & =\frac{1}{3 \sqrt{42}}(8 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+17 \hat{\mathbf{k}}) \end{aligned} \end{array} $

To solve for the expression $2a^2 + 3b^2 + 4c^2$, follow these steps:

Given that $\hat{\mathbf{e}} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}} + c \hat{\mathbf{k}}$ is a unit vector, we have:

$ a^2 + b^2 + c^2 = 1 $

The vector $\hat{\mathbf{e}}$ is coplanar with the vectors $\hat{\mathbf{i}} - 3 \hat{\mathbf{j}} + 5 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}} + \hat{\mathbf{j}} - 5 \hat{\mathbf{k}}$. The condition for coplanarity implies the determinant formed by these vectors should be zero:

$ \left|\begin{array}{ccc} a & b & c \\ 1 & -3 & 5 \\ 3 & 1 & -5 \end{array}\right| = 0 $

Calculating the determinant gives:

$ a(10) - b(-20) + c(10) = 0 \quad \Rightarrow \quad 10a + 20b + 10c = 0 $

Additionally, since $\hat{\mathbf{e}}$ is perpendicular to $\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}$, we have:

$ a + b + c = 0 $

Solving the linear equations, we use the relation derived from coplanarity:

$ 10a + 20b + 10c = 0 \quad \Rightarrow \quad a + 2b + c = 0 $

Together with $ a + b + c = 0 $, it follows:

$ \frac{a}{1} = \frac{-b}{0} = \frac{c}{-1} = \lambda $

Thus:

$ a = \lambda, \quad b = 0, \quad c = -\lambda $

Substituting into the unit vector condition:

$ a^2 + b^2 + c^2 = 1 \quad \Rightarrow \quad \lambda^2 + 0 + (-\lambda)^2 = 1 \quad \Rightarrow \quad 2\lambda^2 = 1 $

Hence:

$ \lambda^2 = \frac{1}{2} \quad \Rightarrow \quad \lambda = \pm \frac{1}{\sqrt{2}} $

Therefore, the values for $a$, $b$, and $c$ are:

$ a = \frac{1}{\sqrt{2}}, \quad b = 0, \quad c = -\frac{1}{\sqrt{2}} $

Calculating $2a^2 + 3b^2 + 4c^2$:

$ 2\left(\frac{1}{2}\right) + 3(0) + 4\left(\frac{1}{2}\right) = 1 + 2 = 3 $

Thus, $2a^2 + 3b^2 + 4c^2 = 3$.

To find the magnitude of vector $\hat{\mathbf{d}}$, which is normal to the plane formed by vectors $\mathbf{a}$ and $\mathbf{b}$, and also satisfies $\hat{\mathbf{d}} \cdot \hat{\mathbf{c}} = 2$, we perform the following calculations:

Firstly, calculate $\mathbf{a} \times \mathbf{b}$, the cross product of vectors $\mathbf{a}$ and $\mathbf{b}$:

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

The calculation proceeds as follows:

$ \begin{aligned} \mathbf{d} = & \ \lambda \left(\hat{\mathbf{i}}(1 \cdot 1 - (-2) \cdot (-2)) - \hat{\mathbf{j}}(1 \cdot 1 - (-2) \cdot 1) + \hat{\mathbf{k}}(1 \cdot (-2) - 1 \cdot 1)\right) \\ = & \ \lambda (\hat{\mathbf{i}}(1 - 4) - \hat{\mathbf{j}}(1 + 2) + \hat{\mathbf{k}}(-2 - 1)) \\ = & \ \lambda (-3\hat{\mathbf{i}} - 3\hat{\mathbf{j}} - 3\hat{\mathbf{k}}) \\ = & \ -3\lambda(\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}) \end{aligned} $

Next, use the condition $\hat{\mathbf{d}} \cdot \hat{\mathbf{c}} = 2$:

$ \mathbf{d} \cdot \mathbf{c} = -3\lambda [2 \hat{\mathbf{i}} + 1 \hat{\mathbf{j}} - 1 \hat{\mathbf{k}}] $

Solves as:

$ \begin{aligned} -3\lambda[2 + 1 - 1] & = 2 \\ -3\lambda \times 2 & = 2 \\ -6\lambda & = 2 \\ \lambda & = -\frac{1}{3} \end{aligned} $

Substitute $\lambda$ back to determine $\mathbf{d}$:

$ \mathbf{d} = \lambda (-3(\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}})) = \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}} $

Finally, the magnitude of $\hat{\mathbf{d}}$ is:

$ |\mathbf{d}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} $

The position vectors and direction vectors for two lines are given:

Line $ L_1 $: $\mathbf{r} = \mathbf{a} + t\mathbf{b}$, where $\mathbf{a} = \hat{\mathbf{i}} - \hat{\mathbf{j}} + 3\hat{\mathbf{k}}$ and $\mathbf{b} = 2\hat{\mathbf{i}} - \hat{\mathbf{j}} + \lambda \hat{\mathbf{k}}$.

Line $ L_2 $: $\mathbf{r} = \mathbf{c} + s\mathbf{d}$, where $\mathbf{c} = -\hat{\mathbf{k}}$ and $\mathbf{d} = \hat{\mathbf{i}} + 2\hat{\mathbf{j}} - \hat{\mathbf{k}}$.

To determine whether the lines are coplanar or skew, we set up equations based on their parametric forms:

$ L_1 = \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix} + t\begin{pmatrix} 2 \\ -1 \\ \lambda \end{pmatrix} $

$ L_2 = \begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix} + s\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} $

By equating the components of the parametric equations, we get:

$ 1 + 2t = s $

$-1 - t = 2s$

$3 + \lambda t = -1 - s$

Solving the first two equations simultaneously gives:

$ 2 + 4t = 2s $

$ -1 - t = 2s $

$ \frac{-1 - t}{2s} = \frac{3 + 5t}{0} $

From these equations, solve for $ t $ and $ s $:

$\Rightarrow t = -\frac{3}{5}$

$\Rightarrow s = -\frac{1}{5}$ after substituting $ t $.

For the third equation:

$ 3 - \frac{3\lambda}{5} = -1 + \frac{1}{5} $

From this, solve for $\lambda$:

$ 21 - \frac{1}{5} = \frac{3\lambda}{5} $

$ \Rightarrow \lambda = \frac{19}{3} $

Thus, the lines are skew when $\lambda \neq \frac{19}{3}$. If $\lambda = \frac{19}{3}$, the lines would be coplanar.

Given that the vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ each have a magnitude of $\sqrt{2}$ and the angle between any two vectors is $\frac{\pi}{3}$, we can analyze the vectors $\mathbf{x}$ and $\mathbf{y}$ defined as follows:

$ |\mathbf{a}| = |\mathbf{b}| = |\mathbf{c}| = \sqrt{2} $

First, let's express $\mathbf{x}$:

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

Using the vector triple product identity:

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

Given that the angle between each pair of vectors is $\frac{\pi}{3}$, the dot products are:

$ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos\left(\frac{\pi}{3}\right) = \sqrt{2} \times \sqrt{2} \times \frac{1}{2} = 1 $

Therefore, the expression for $\mathbf{x}$ becomes:

$ \mathbf{x} = 1 \cdot \mathbf{b} - 1 \cdot \mathbf{c} = \mathbf{b} - \mathbf{c} $

Now, consider $\mathbf{y}$:

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

Using the vector triple product identity again:

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

This simplifies similarly to:

$ \mathbf{y} = 1 \cdot \mathbf{c} - 1 \cdot \mathbf{a} = \mathbf{c} - \mathbf{a} $

Hence, the magnitudes of $\mathbf{x}$ and $\mathbf{y}$ are:

$ |\mathbf{x}| = |\mathbf{b} - \mathbf{c}| $

$ |\mathbf{y}| = |\mathbf{c} - \mathbf{a}| $

Since both expressions have the same form, it follows that:

$ |\mathbf{x}| = |\mathbf{y}| $

Given that the vector $\mathbf{a}$ is perpendicular to the plane formed by the non-zero vectors $\mathbf{b}$ and $\mathbf{c}$, we have:

$\mathbf{a} \cdot \mathbf{b} = 0$

$\mathbf{a} \cdot \mathbf{c} = 0$

From the problem statement, we know:

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

Expanding both sides, we get:

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

This simplifies to:

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

Hence, we conclude:

$ \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a} = 0 $

Now, considering $|(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}| + |(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|$, we find:

$ \left|\left(|\mathbf{a}||\mathbf{b}| \sin 90^\circ \right) \cdot \mathbf{c}\right| + |(\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{b} \cdot \mathbf{c}) \mathbf{a}| $

Since $\sin 90^\circ = 1$, this becomes:

$ |\mathbf{a}||\mathbf{b}||\mathbf{c}| + 0 - 0 \quad \text{(because $\mathbf{a} \cdot \mathbf{c} = 0$ and $\mathbf{b} \cdot \mathbf{c} = 0$)} $

Thus, the result is:

$ |\mathbf{a}||\mathbf{b}||\mathbf{c}| $

Given the relationships $ \mathbf{a} \times \mathbf{c} = \mathbf{b} $ and $ \mathbf{a} \cdot \mathbf{c} = 3 $, we need to find $ \mathbf{a} \cdot (\mathbf{c} \times \mathbf{b} - \mathbf{b} - \mathbf{c}) $.

First, consider:

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

We are given $ |\mathbf{b}|^2 = (\mathbf{a} \times \mathbf{c})^2 $. Thus:

$ |\mathbf{b}|^2 + 9 = |\mathbf{a}|^2 \cdot |\mathbf{c}|^2 $

Start with:

$ 27 + 9 = 6 \cdot |\mathbf{c}|^2 \implies 36 = 6 \cdot |\mathbf{c}|^2 \implies |\mathbf{c}|^2 = 6 $

We then know:

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

Substituting:

$ 6\mathbf{a} - 3\mathbf{c} = \mathbf{c} \times \mathbf{b} $

Now compute:

$ \mathbf{a} \cdot (6 \mathbf{a} - 3 \mathbf{c} - \mathbf{b} - \mathbf{c}) $

Breaking it down:

$ 6 \mathbf{a}^2 = 6 \times |\mathbf{a}|^2 = 6 \times 6 $

$ -4 (\mathbf{a} \cdot \mathbf{c}) = -4 \times 3 = -12 $

$ \mathbf{a} \cdot \mathbf{b} = 0 $ because $ \mathbf{a} \times \mathbf{c} = \mathbf{b} $ implies orthogonality of $ \mathbf{a} $ and $ \mathbf{b} $.

Substituting:

$ = 6 \times 6 - 12 - 0 = 36 - 12 - 0 = 24 $

Thus, the final result is:

$ 24 $

Position vector of

$ P=\frac{4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-6 \hat{\mathbf{k}}+4 \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}}{3}=\frac{8 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}}{3} $

Position vector of

$ \begin{aligned} Q & =\frac{8 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-12 \hat{\mathbf{k}}+2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{3} \\\\ & =\frac{10 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-9 \hat{\mathbf{k}}}{3} \end{aligned} $

Let the point which divides $ P Q $ in ratio $ 2: 3 $ be $ M $.

Position vector of $ M $

$ \begin{array}{l} =\frac{2\left(\frac{10 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-9 \hat{\mathbf{k}}}{3}\right)+3\left(\frac{8 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}}{3}\right)}{5} \\\\ =\frac{20 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-18 \hat{\mathbf{k}}+24 \hat{\mathbf{i}}-27 \hat{\mathbf{j}}}{15} \\\\ =\frac{44 \hat{\mathbf{i}}-33 \hat{\mathbf{j}}-18 \hat{\mathbf{k}}}{15} \end{array} $

Line joining the points ( $ 1,-1,1 $ ) and $ (1,1,-2) $ be

$ \frac{x-1}{0}=\frac{y+1}{2}=\frac{z-1}{-3} $

Line joining the points $ (2,1,-6) $ and $ (3,-1,-7) $ is

$ \frac{x-2}{1}=\frac{y-1}{-2}=\frac{z+6}{-1} $

For intersection of points,

$ \frac{x-1}{0}=\frac{y+1}{2}=\frac{z-1}{-3}=K $

$ \Rightarrow x=1, y=2 K-1 \text { and } Z=-3 K+1 $

$ \text { and } \frac{x-2}{1}=\frac{y-1}{-2}=\frac{z+6}{-1}=\lambda $

$ x=\lambda+2, y=1-2 \lambda, z=-\lambda-6 $

$ \therefore \lambda=-1 $ and $ K=2 $

$ \therefore $ Positivion vector of points of intersection is $ \mathbf{i}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} $

$ \mathbf{a}=4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} $ and $ \mathbf{b}=6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} $

$ \begin{array}{l} \text { Required magnitude }=\frac{\mathbf{a} \cdot \mathbf{b}}{|a|} \\\\ \qquad \begin{aligned} & =\frac{24-10+6}{\sqrt{16+25+9}}=\frac{20}{\sqrt{50}}=\frac{20}{5 \sqrt{2}} \\\\ & =2 \sqrt{2} \end{aligned} \end{array} $

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

Given, $ |\mathbf{d}|=1 \Rightarrow x^{2}+y^{2}+z^{2}=1 \quad \ldots \text { (i) } $

$ d \cdot \mathbf{c}=0 \Rightarrow-x+2 z=0 \Rightarrow x=2 z \quad \ldots \text { (ii) } $

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

$ \left|\begin{array}{ccc}2 & -1 & 0 \\\\ 0 & 2 & -1 \\\\ x & y & z\end{array}\right|=0 $

$ \begin{array}{l} 2(2 z+y)+x(l)=0 \\\\ 4 z+2 y+2 z=0 \\\\ 6 z+2 y=0 \Rightarrow y=-3 z \end{array} $

$ [\because x=2 z] $

From Eq. (i), $ z^{2}+4 z^{2}+9 z^{2}=1 $

$ \begin{aligned} 14 z^{2} & =1 \\\\ z^{2} & =\frac{1}{14} \Rightarrow z=\frac{ \pm 1}{\sqrt{14}} \end{aligned} $

If $ z=\frac{1}{\sqrt{14}}, x=\frac{2}{\sqrt{14}}, y=\frac{-3}{\sqrt{14}} $

If $ z=\frac{-1}{\sqrt{14}}, x=\frac{-2}{\sqrt{14}}, y=\frac{3}{\sqrt{14}} $

$ |d \cdot b|=\left(\frac{6}{\sqrt{14}}+\frac{1}{\sqrt{14}}\right)=\frac{7}{\sqrt{14}}=\sqrt{\frac{7}{2}} $

Given,

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non-coplanar vectors and the three points $\lambda \mathbf{a}-2 \mathbf{b}+\mathbf{c}, 2 \mathbf{a}+\lambda \mathbf{b}-2 \mathbf{c}$, $4 a+7 b-8 c$ are collinear.

$ \begin{aligned} \text { Let } P & =\lambda \mathbf{a}-2 \mathbf{b}+\mathbf{c} \\\\ Q & =2 \mathbf{a}+\lambda \mathbf{b}-2 \mathbf{c} \\\\ R & =4 \mathbf{a}+7 \mathbf{b}-8 \mathbf{c} \\\\ \mathbf{P Q} & =(2 \mathbf{a}+\lambda \mathbf{b}-2 \mathbf{c})-(\lambda \mathbf{a}-2 \mathbf{b}+\mathbf{c}) \\\\ & =(2-\lambda) \mathbf{a}+(\lambda+2) \mathbf{b}+(-3) \mathbf{c} \\\\ \mathbf{P R} & =(4 a+7 b-8 \mathbf{c})-(\lambda \mathbf{a}-2 \mathbf{b}+\mathbf{c}) \\\\ & =(4-\lambda) \mathbf{a}+9 \mathbf{b}-9 \mathbf{c} \end{aligned} $

$\because P, Q$ and $R$ collinear.

$ \begin{aligned} & \therefore \mathbf{P Q}=k \mathbf{P R} \\\\ & \begin{array}{c} \Rightarrow(2-\lambda) \mathbf{a}+(\lambda+2) \mathbf{b}+(-3) \mathbf{c} \\\\ \quad=k(4-\lambda) \mathbf{a}+9 k \mathbf{b}-9 k \mathbf{c} \\\\ \therefore \quad 2-\lambda=k(4-\lambda) \Rightarrow \lambda+2=9 k \\\\ \quad-3=-9 k \Rightarrow k=\frac{1}{3} \end{array} \end{aligned} $

Now, $\lambda+2=9\left(\frac{1}{3}\right)=3$

$ \begin{array}{lrl} \Rightarrow & 6-3 \lambda & =4-\lambda \Rightarrow 2=2 \lambda \\\\ \Rightarrow & & \lambda=1 \end{array} $

We have,

$ \begin{aligned} |\mathbf{a}| & =3,|\mathbf{b}|=4 \\\\ |\mathbf{a}+\mathbf{b}| & =\sqrt{37},|\mathbf{a}-\mathbf{b}|=k \text { and } \\\\ (\mathbf{a}, \mathbf{b}) & =0 \end{aligned} $

Now, $|\mathbf{a}+\mathbf{b}|=\sqrt{37}$

$ \begin{array}{rr} \Rightarrow & (\mathbf{a}+\mathbf{b}) \cdot(\mathbf{a}+\mathbf{b})=37 \\\\ \Rightarrow & |\mathbf{a}|^2+|\mathbf{b}|^2+2 \mathbf{a} \cdot \mathbf{b}=37 \\\\ \Rightarrow & 9+16+2|\mathbf{a}||\mathbf{b}| \cos \theta=37 \\\\ \Rightarrow & 2(3)(4) \cos \theta=12 \\\\ \Rightarrow & \cos \theta=\frac{1}{2} \\\\ \Rightarrow & \theta=60^{\circ} \end{array} $

Now, $|\mathbf{a}-\mathbf{b}|=k$

$ \begin{array}{lr} \Rightarrow (\mathbf{a}-\mathbf{b}) \cdot(\mathbf{a}-\mathbf{b})=k^2 \\\\ \Rightarrow|\mathbf{a}|^2+|\mathbf{b}|^2-2|\mathbf{a}||\mathbf{b}| \cos (\theta)=k^2 \\\\ \Rightarrow 9+16-2(3)(4) \cos \left(60^{\circ}\right)=k^2 \end{array} $

$ \begin{aligned} & \Rightarrow 25-24\left(\frac{1}{2}\right)=k^2 \\\\ & \Rightarrow k^2=13 \\\\ & \therefore \frac{4}{13}(k \sin \theta)^2 \\\\ & =\frac{4}{13} k^2 \sin ^2\left(60^{\circ}\right) \\\\ & =\frac{4}{13}(13)\left(\frac{\sqrt{3}}{2}\right)^2=3 \end{aligned} $

Given,

$\mathbf{r}$ is a vector perpendicular to the plane determined by the vectors $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}$ and $\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$

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

Also, given that

The magnitude of the projection of $\mathbf{r}$ on the vector $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ is 1

$ \begin{aligned} & \therefore \quad\left|\frac{\mathbf{r} \cdot(2 \hat{i}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})}{\sqrt{2^2+1^2+2^2}}\right|=1 \\\\ & \Rightarrow \quad|\lambda(-4-4+4)|=3 \Rightarrow \lambda=\frac{3}{4} \\\\ & \therefore \quad \mathbf{r}=\frac{3}{4}(-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} $

Now,

$ \begin{aligned} |r| & =\frac{3}{4} \sqrt{(-2)^2+(4)^2+(2)^2}=\frac{3}{4} \sqrt{24} \\\\ & =\frac{3}{4}(2 \sqrt{6})=\frac{3 \sqrt{6}}{2} \end{aligned} $

Given,

$ \begin{aligned} & \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\\\ & \cos (\mathbf{a}, \mathbf{b} \times \mathbf{c})=\sqrt{\frac{2}{3}} \text { and } \mathbf{a} \text { is a unit vectors. } \\\\ & \text { Now, } \mathbf{b} \times \mathbf{c}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\\\ 1 & -1 & 2 \\\\ 1 & 2 & -1 \end{array}\right|=-3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\\\ & \therefore|\mathbf{a} \times(\mathbf{b} \times \mathbf{c})| \\\\ & =|\mathbf{a}||\mathbf{b} \times \mathbf{c}| \sin (\mathbf{a}, \mathbf{b} \times \mathbf{c}) \\\\ & =(1) \sqrt{(-3)^2+(3)^2+(3)^2} \sqrt{1-(\cos (\mathbf{a}, \mathbf{b} \times \mathbf{c}))^2} \\\\ & =3 \sqrt{3} \sqrt{1-\left(\sqrt{\frac{2}{3}}\right)^2}=3 \sqrt{3}\left(\frac{1}{\sqrt{3}}\right)=3 \end{aligned} $

$A B=\sqrt{(4-3)^2+(1-2)^2+(0+1)^2}$

$ \begin{aligned} & =\sqrt{1+1+1}=\sqrt{3} \\\\ A C & =\sqrt{(2-3)^2+(1-2)^2+(4+1)^2} \\\\ & =\sqrt{1+1+25} \\\\ & =\sqrt{27}=3 \sqrt{3} \end{aligned} $

By angle bisector theorem,

$ \frac{B D}{C D}=\frac{A B}{A C}=\frac{\sqrt{3}}{3 \sqrt{3}}=\frac{1}{3}=1: 3 $

The point $D$ divides the segment $B C$ internally in the ratio $1: 3$

$\therefore$ Coordinate of $D$ will be

$ \left(\frac{4(3)+2(1)}{1+3}, \frac{1(3)+1(1)}{1+3}, \frac{0(3)+4(1)}{1+3}\right) $

i.e. $\left(\frac{7}{2}, 1,1\right)$

Given, $D \equiv(p, q, r)$

$\therefore \quad p=\frac{7}{2}, q=1$ and $r=1$

Now, $\sqrt{2 p+q+r}=\sqrt{2\left(\frac{7}{2}\right)+1+1}=\sqrt{9}=3$

Given, $(3,0,2)$ and $(0,2, k)$ are the direction ratios of two lines and $\theta$ is the angle between them.

$ \begin{aligned} & \therefore \cos \theta=\frac{3 \times 0+0 \times 2+2 \times k}{\sqrt{3^2+0^2+2^2} \cdot \sqrt{0^2+2^2+k^2}} \\\\ & \Rightarrow|\cos \theta|=\left|\frac{2 k}{\sqrt{13\left(4+k^2\right)}}\right| \\\\ & \Rightarrow\left|\frac{2 k}{\sqrt{13\left(4+k^2\right)}}\right|=\frac{6}{13}\left[\text { given, }|\cos \theta|=\frac{6}{13}\right] \\\\ & \Rightarrow \quad \frac{4(k)^2}{13\left(4+k^2\right)}=\left(\frac{6}{13}\right)^2 \Rightarrow \frac{4 k^2}{4+k^2}=\frac{36}{13} \\\\ & \Rightarrow \quad 52 k^2=144+36 k^2 \\\\ & \Rightarrow \quad 16 k^2=144 \Rightarrow k^2=9 \\\\ & \Rightarrow \quad k= \pm 3 \end{aligned} $

We have,

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

$ B(2 \hat{i}-\hat{j}+\hat{k}) \quad C(\hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) $

Position vector of

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

Position vector of

$ \begin{aligned} E & =\frac{\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}+\hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{2} \\\\ & =\frac{1}{2}(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\\\ \therefore \mathrm{DE}= & \frac{1}{2}(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{k}})-\frac{3}{2}(\hat{\mathbf{i}}-\hat{\mathbf{k}}) \\\\ = & -\frac{1}{2} \hat{\mathbf{i}}-\frac{3}{2} \hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} $

$\Rightarrow$ Unit vector along

$\begin{aligned} \mathrm{DE} & =\frac{\left(-\frac{1}{2} \hat{\mathbf{i}}-\frac{3}{2} \hat{\mathbf{j}}+\hat{\mathbf{k}}\right)}{\sqrt{\left(-\frac{1}{2}\right)^2+\left(\frac{-3}{2}\right)^2+(1)^2}} \\ & =\frac{1}{\sqrt{14}}(-\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\end{aligned}$

$ \begin{aligned} & \text { Let } \mathbf{a}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\\\ & \mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\\\ & \therefore \quad|a|=3,|b|=3 \end{aligned} $

A vector along bisectors $=\frac{\mathbf{a}}{|\mathbf{a}|} \pm \frac{b}{|b|}$

$ \begin{aligned} & =\frac{2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}}{3} \pm \frac{\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{3} \\\\ & =\frac{3 \hat{\mathbf{i}}+3 \hat{\mathbf{k}}}{3}, \frac{\hat{\mathbf{i}}-4 \hat{\mathbf{j}}-\hat{\mathbf{k}}}{3} \end{aligned} $

$\begin{aligned} \therefore \text { Required vector }=\hat{\mathbf{i}} & +\hat{\mathbf{k}} \\\\ & (\because \text { Magnitude }=\sqrt{2})\end{aligned}$

$ \begin{aligned} & \cos \theta=\frac{(4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})}{\sqrt{4^2+(-1)^2+2^2} \cdot \sqrt{1^2+3^2+\left(-2^2\right.}} \\\\ & \begin{aligned} \cos \theta & =\frac{4-3-4}{\sqrt{21} \cdot \sqrt{14}}=\frac{-3}{7 \sqrt{6}} \Rightarrow \sin \theta=\sqrt{\frac{285}{294}} \\\\ \text { So, } \sin 2 \theta & =2 \sin \theta \cos \theta \\\\ & =2 \cdot \frac{\sqrt{285}}{7 \sqrt{6}}\left(\frac{-3}{7 \sqrt{6}}\right)=\frac{-\sqrt{285}}{49} \end{aligned} \end{aligned} $

We have, $|a|=3,|b|=2 \sqrt{2}$ and $|c|=5$

$\because \mathrm{c}$ is perpendicular to the plane of a and b .

$ \begin{aligned} & \begin{aligned} & \therefore \quad k(\mathbf{a} \times \mathbf{b}) \Rightarrow \mathbf{c} \perp \mathbf{a}, \mathbf{c} \perp \mathbf{b} \\\\ & \text { Now, }|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=|\mathbf{a}|^2+|b|^2+|c|^2+2 \end{aligned} \\\\ & \quad(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \\\\ & =9+8+25+2\left(3 \times 2 \sqrt{2} \cdot \sin \frac{\pi}{4}+0+0\right) \\\\ & |\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=42+12=54 \\\\ & \Rightarrow|\mathbf{a}+\mathbf{b}+\mathbf{c}|=3 \sqrt{6} \end{aligned} $

Let $A(\lambda \mathbf{a}+3 \mathbf{b}-\mathbf{c}), B(\mathbf{a}-\lambda \mathbf{b}+3 \mathbf{c})$

$ \begin{aligned} & \mathbf{c}(3 \mathbf{a}+4 \mathbf{b}-\lambda \mathbf{c}) \text { and } D(\mathbf{a}-6 \mathbf{b}+6 \mathbf{c}) \\\\ & \mathbf{A D}=(1-\lambda) \mathbf{a}-9 \mathbf{b}+7 \mathbf{c} \\\\ & \mathbf{B D}=(\lambda-6) \mathbf{b}+3 \mathbf{c} \\\\ & \mathbf{C D}=-2 \mathbf{a}-10 \mathbf{b}+(6+\lambda) \mathbf{c} \end{aligned} $

$\because A, B, C$ and $D$ are coplanar

$ \Rightarrow\left|\begin{array}{ccc} 1-\lambda & -9 & 7 \\\\ 0 & \lambda-6 & 3 \\\\ -2 & -10 & 6+\lambda \end{array}\right|=0 $

$\begin{aligned} & \Rightarrow \quad(1-\lambda)\left[\lambda^2-36+30\right] \\\\ & -2[-27-7 \lambda+42]=0 \\\\ & \Rightarrow \quad(1-\lambda)\left(\lambda^2-6\right)-2(15-7 \lambda)=0 \\\\ & \Rightarrow \quad \lambda^2-6-\lambda^3+6 \lambda-30+14 \lambda=0 \\\\ & \Rightarrow \quad-\lambda^3+\lambda^2+20 \lambda-36=0 \\\\ & \Rightarrow \quad \lambda^3-\lambda^2-20 \lambda+36=0 \\\\ & \Rightarrow \quad(\lambda-2)\left(\lambda^2+\lambda-18\right)=0 \Rightarrow \lambda=2\end{aligned}$