Vector Algebra

$ \begin{aligned} & \text { Here } \overrightarrow{\mathrm{AC}}=(-2-2) \hat{i}+(-3-1) \hat{j}+(5-1) \hat{k} \\\\ & =-4 \hat{i}-4 \hat{j}+4 \hat{k} \\\\ & \overrightarrow{\mathrm{BD}}=(1-1) \hat{i}+(-6-2) \hat{j}+(-7-5) \hat{k} \\\\ & =-8 \hat{j}-12 \hat{k} \end{aligned} $

So, area of quadrilateral $=\frac{1}{2}| \overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}} \mid$

$ \begin{aligned} & =\frac{1}{2}\left\|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -4 & -4 & 4 \\ 0 & -8 & -12 \end{array}\right\| \\\\ & =\frac{1}{2}|(48+32) \hat{i}-(48-0) \hat{j}+(32-0) \hat{k}| \\\\ & =\frac{1}{2}|80 \hat{i}-48 \hat{j}+32 \hat{k}| \\\\ & =\frac{1}{2} 16|15 \hat{i}-3 \hat{j}+2 \hat{k}| \\\\ & =8 \sqrt{25+9+4}=8 \sqrt{38} \text { sq units. } \end{aligned} $

Given : Points with position vectors

$ \alpha \hat{i}+10 \hat{j}+13 \hat{k}, 6 \hat{i}+11 \hat{j}+11 \hat{k} $

and $\frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k}$ are collinear.

So, $\frac{\alpha-6}{6-\frac{9}{2}}=\frac{10-11}{11-\beta}=\frac{13-11}{11+8}$

$ \begin{aligned} & \Rightarrow \frac{2(\alpha-6)}{3}=\frac{-1}{11-\beta}=\frac{2}{19} \\\\ & \Rightarrow \frac{2}{3}(\alpha-6)=\frac{2}{19} \end{aligned} $

$ \begin{aligned} & \Rightarrow 19 \alpha-114=3 \Rightarrow 19 \alpha=117 \\\\ & \Rightarrow \alpha=\frac{117}{19} \end{aligned} $

And, $\frac{-1}{11-\beta}=\frac{2}{19}$

$ \begin{aligned} & \Rightarrow-19=22-2 \beta \\\\ & \Rightarrow 2 \beta=41 \\\\ & \Rightarrow \beta=\frac{41}{2} \end{aligned} $

$ \begin{aligned} & \therefore(19 \alpha-6 \beta)^2=\left(19 \times \frac{117}{19}-\frac{6 \times 41}{2}\right)^2 \\\\ & =(117-123)^2=36 \end{aligned} $

Given that the volume $V$ of the parallelepiped formed by the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is represented by the scalar triple product $[\vec{a},\vec{b},\vec{c}]$, which is the determinant of the 3 x 3 matrix with vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ as its rows (or columns).

When the vectors representing the coterminous edges of the parallelepiped are $\vec{a}$, $\vec{b} + \vec{c}$, and $\vec{a} + 2\vec{b} + 3\vec{c}$, the volume $V$ of the parallelepiped is represented by :

$\begin{aligned} & V=[\vec{a}, \vec{b}+\vec{c}, \vec{a}+2 \vec{b}+3 \vec{c}] \\\\ & =\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 3\end{array}\right|\left[\begin{array}{ll}\vec{a} & \vec{b} & \vec{c}\end{array}\right] \\\\ & =1(3-2)[\vec{a}, \vec{b}, \vec{c}] \\\\ & =\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=V \end{aligned}$

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

$ \begin{aligned} & \overrightarrow{O B}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\\\ & \overrightarrow{O C}=(a+1) \hat{\mathbf{i}}+2 \hat{\mathbf{k}} \end{aligned} $

and $ \overrightarrow{O D}=9 \hat{\mathbf{i}}+(a-8) \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$

$ \begin{array}{ll} &\therefore \overrightarrow{A B}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} \\\\ & \overrightarrow{A C}=a \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\\\ &\text { and } \overrightarrow{A D}=8 \hat{\mathbf{i}}+(a-6) \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{array} $

Since, given point are coplanar

$ \begin{aligned} & \therefore \quad[\overrightarrow{A B}, \overrightarrow{A C}, \overrightarrow{A D}]=0 \\\\ & \Rightarrow\left|\begin{array}{ccr} 1 & -1 & 1 \\ a & 2 & -1 \\ 8 & a-6 & 3 \end{array}\right|=0 \\\\ & \Rightarrow 1(6+a-6)+1(3 a+8)+1\left(a^2-6 a-16\right)=0 \\\\ & \Rightarrow a+3 a+8+a^2-6 a-16=0 \Rightarrow a^2-2 a-8=0 \\\\ & \Rightarrow a^2-4 a+2 a-8=0 \Rightarrow a(a-4)+2(a-4)=0 \\\\ & \Rightarrow(a-4)(a+2)=0 \Rightarrow a=4,-2 \\\\ & \therefore \text { Sum of all values of } a=4-2=2 \end{aligned} $

Given, position vectors of the points $A, B, C$ and $D$ be

$5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}, \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}},-2 \hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$

$ \begin{aligned} \overrightarrow{A B} & =(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}})=-4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+(3-2 \lambda) \hat{\mathbf{k}} \\\\ \overrightarrow{A C} & =(-2 \hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}+4 \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}) \\\\ & =-7 \hat{\mathbf{i}}+(\lambda-5) \hat{\mathbf{j}}+(4-2 \lambda) \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} \text { and } \overrightarrow{A D} & =(-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}) \\\\ & =-6 \hat{\mathbf{i}}+(6-2 \lambda) \hat{\mathbf{k}} \end{aligned} $

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

$ \begin{aligned} & \therefore [ { \overrightarrow{AB}~ \overrightarrow{AC}~ \overrightarrow{AD}] }=0 \\\\ & \Rightarrow \left|\begin{array}{ccc} -4 & -3 & (3-2 \lambda) \\ -7 & (\lambda-5) & (4-2 \lambda) \\ -6 & 0 & 6-2 \lambda \end{array}\right|=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow -6\left\{(-12+6 \lambda)-\left(13 \lambda-15-2 \lambda^2\right)\right\} \\\\ & +(6-2 \lambda)\{-4 \lambda+20-21\} =0 \\\\ & \Rightarrow -6\left(2 \lambda^2-7 \lambda+3\right)+(6-2 \lambda)(-4 \lambda-1) =0 \\\\ & \Rightarrow -12 \lambda^2+42 \lambda-18+8 \lambda^2-22 \lambda-6 =0 \\\\ & \Rightarrow -4 \lambda^2+20 \lambda-24 =0 \\\\ & \Rightarrow \lambda^2-5 \lambda+6 =0 \\\\ & \Rightarrow (\lambda-2)(\lambda-3) =0 \\\\ & \Rightarrow \lambda =2,3 \end{aligned} $

$ \therefore \sum\limits_{\lambda \varepsilon S}(\lambda+2)^2=(2+2)^2+(3+2)^2=16+25=41 $

Given vectors :

$ \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} $

$ \vec{b} = \hat{i} - 2\hat{j} - 2\hat{k} $

$ \vec{c} = -\hat{i} + 4\hat{j} + 3\hat{k} $

Since $ \vec{d} $ is perpendicular to both $ \vec{b} $ and $ \vec{c} $, its direction is given by their cross product :

$ \vec{d} = \lambda(\vec{b} \times \vec{c}) $

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

Thus, $ \vec{b} \times \vec{c} = 2\hat{i} - \hat{j} + 2\hat{k} $

Given this, $ \vec{d} $ can be expressed as :

$ \vec{d} = \lambda(2\hat{i} - \hat{j} + 2\hat{k}) $

Using the given condition that $ \vec{a} \cdot \vec{d} = 18 $ :

$ (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot \lambda(2\hat{i} - \hat{j} + 2\hat{k}) = 18 $

$ \Rightarrow 2\lambda(2) - 3\lambda(1) + 4\lambda(2) = 18 $

$ \Rightarrow \lambda(4 + 8 - 3) = 18 $

$ \Rightarrow 9\lambda = 18 $

$ \Rightarrow \lambda = 2 $

So, $ \vec{d} = 4\hat{i} - 2\hat{j} + 4\hat{k} $

Now, using the identity :

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

Given $ |\vec{a}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29} $

and $ |\vec{d}| = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$

Using your corrected calculations :

$ |\vec{a} \times \vec{d}|^2 = 29 \times 36 - 18^2 = 1044 - 324 = 720 $

$\begin{aligned} & \text { Projection of }\vec{a} \text { on } \vec{b} =\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \\\\ & = \frac{(5 \hat{i}-\hat{j}-3 \hat{k}) \cdot(\hat{i}+3 \hat{j}+5 \hat{k})}{\sqrt{1^2+3^2+5^2}}=\frac{5-3-15}{\sqrt{35}} \\\\ & = \frac{-13}{\sqrt{35}}\end{aligned}$

Negative sign indicates that direction of the projection vector is opposite to the direction of $\vec{b}$.

$\begin{aligned} & \vec{r} \times \vec{a}=\vec{c} \times \vec{a} \\\\ & \Rightarrow(\vec{r}-\vec{c}) \times \vec{a}=0 \Rightarrow \vec{r}-\vec{c}=\lambda \vec{a}((\vec{r}-\vec{c} ) \text{and} \overrightarrow{a} \text { are parallel }) \\\\ & \Rightarrow \vec{r}=\vec{c}+\lambda \vec{a} \\\\ & \Rightarrow \vec{r} \cdot \vec{b}=\vec{c} \cdot \vec{b}+\lambda \vec{a} \cdot \vec{b} \\\\ & 0=(1-3)+\lambda(2+5) \Rightarrow \lambda=\frac{2}{7} \\\\ & \text { Hence, } \vec{r}=\vec{c}+\frac{2 \vec{a}}{7} \\\\ & \vec{r} \Rightarrow \frac{11}{7} \hat{i}-\frac{11}{7} \hat{k} \\\\ & |\vec{r}|=\sqrt{\left(\frac{11}{7}\right)^2+\left(-\frac{11}{7}\right)^2} \Rightarrow r=\frac{11 \sqrt{2}}{7}\end{aligned}$

$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{c}}=\hat{5 \mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}) \times \vec{b}=0, \quad \vec{r} \cdot \vec{a}=0$

$\Rightarrow \vec{r}-\vec{c}=\lambda \vec{b}$

Also, $(\vec{c}+\lambda \vec{b}) \cdot \vec{a}=0$

$\Rightarrow \vec{a} \cdot \vec{c}+\lambda(\vec{a} \cdot \vec{b})=0$

$\therefore \lambda=\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}=\frac{-8}{5}$

$\overrightarrow{\mathrm{r}}=\frac{5(5 \hat{\mathrm{i}}-3 \hat{\mathrm{i}}+3 \hat{\mathrm{k}})-8(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{5}$

$ \Rightarrow $ $\overrightarrow{\mathrm{r}}=\frac{17 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{5}$

$ \Rightarrow $ $|\overrightarrow{\mathrm{r}}|^{2}=\frac{1}{25}(289+50)$

$ \Rightarrow $ $25|\overrightarrow{\mathrm{r}}|^{2}=339$

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

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

$ \begin{aligned} & \Rightarrow |\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \vec{a}) \\\\ & =|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{c}-\vec{c} \cdot \vec{a}) \\\\ & \Rightarrow \vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=0 \Rightarrow \vec{c} \cdot \vec{a}=0 \\\\ & |\vec{a}+\lambda \vec{c}|^{2}=|\vec{a}|^{2}+\lambda^{2}|\vec{c}|^{2}+0 \geq|\vec{a}|^{2} \end{aligned} $

So $\mathrm{A}$ is correct.

$B$ is incorrect.

$\left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right) + \overrightarrow b \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right) \times \left( {\overrightarrow a - \overrightarrow b } \right)$

$ = \left( {\overrightarrow a \left( {\overrightarrow a .\,\overrightarrow b } \right) - \overrightarrow b \left( {\overrightarrow a .\,\overrightarrow a } \right) + \overrightarrow a \left( {\overrightarrow b .\,\overrightarrow b } \right) - \overrightarrow b \left( {\overrightarrow a .\,\overrightarrow b } \right)} \right) \times \left( {\overrightarrow a - \overrightarrow b } \right)$

$ = \left( {\overrightarrow a .\,\overrightarrow b } \right)\left( {\overrightarrow a \times \overrightarrow a - \overrightarrow a \times \overrightarrow b } \right) - \left( {\overrightarrow a .\,\overrightarrow a } \right)\left( {\overrightarrow b \times \overrightarrow a - \overrightarrow b \times \overrightarrow b } \right) + \left( {\overrightarrow b .\,\overrightarrow b } \right)\left( {\overrightarrow a \times \overrightarrow a - \overrightarrow a \times \overrightarrow b } \right) - \left( {\overrightarrow a .\,\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a - \overrightarrow b \times \overrightarrow b } \right)$

$ = \left( {\overrightarrow a .\,\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right) - \left( {\overrightarrow a .\,\overrightarrow a } \right)\left( {\overrightarrow b \times \overrightarrow a } \right) + \left( {\overrightarrow b .\,\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right) - \left( {\overrightarrow a .\,\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right)$

$ = \left( {\overrightarrow b \times \overrightarrow a } \right)\left( {\overrightarrow b .\,\overrightarrow b - \overrightarrow a .\,\overrightarrow a } \right)$

$ = \left( {5\overrightarrow b \times \overrightarrow a } \right)\left( {5 + {\lambda ^2} - 13 - {\lambda ^2}} \right)$

$ = 8\left( {\overrightarrow a \times \overrightarrow b } \right)$

$\overrightarrow a \times \overrightarrow b = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr \lambda & 2 & { - 3} \cr 1 & { - \lambda } & 2 \cr } } \right|$

$ = \widehat i(4 - 3\lambda ) - \widehat j(2\lambda + 3) + \widehat k( - {\lambda ^2} - 2)$

$ \Rightarrow \lambda = 1$

${\left| {\overrightarrow a \times \left( {\overrightarrow a - \overrightarrow b } \right) + \overrightarrow b \times \left( {\overrightarrow a - \overrightarrow b } \right)} \right|^2}$

$ = {\left| {2\left( {\overrightarrow a \times \overrightarrow b } \right)} \right|^2} = 4\,.\,35 = 140$

$\overrightarrow b .\overrightarrow c = \overrightarrow b .(2\overline a \times \overrightarrow b ) - 3\overline b .\overrightarrow b $

$ = 0 - 3|\overline b {|^2} = - 48$

$\widehat n = \alpha \overrightarrow b - \overrightarrow a $

$\overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right) = \left( {\overrightarrow c \,.\,\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow c \,.\,\overrightarrow a } \right)\overrightarrow b $

$ = 12\overrightarrow a - \left( {\overrightarrow c \,.\,\left( {\alpha \overrightarrow b - \widehat n} \right)} \right)\overrightarrow b $

$ = 12\overrightarrow a - (12\alpha - 0)\overrightarrow b $

$ = 12\left( {\overrightarrow a - \alpha \overrightarrow b } \right)$

$\therefore$ $\left| {\overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right| = 12$

Let $A \equiv (1,2,3),B \equiv (2,3,4),C \equiv (1,5,7)$

$\overrightarrow n = \overrightarrow {AB} \times \overrightarrow {AC} = \left| {\matrix{ i & j & k \cr 1 & 1 & 1 \cr 0 & 3 & 4 \cr } } \right|$

$ = \widehat i - 4\widehat j + 3\widehat k$

$\widehat {OP} = {{ \pm (\widehat i - 4\widehat j + 3\widehat k)} \over {\sqrt {26} }}$

Since $\cos \beta > 0$, take $-$ sign

$\widehat {OP} = {{\widehat i - 4\widehat j + 3\widehat k} \over {\sqrt {26} }}$

$ \Rightarrow \cos \alpha < 0,\cos \gamma < 0$

$\alpha ,\gamma \in \left( {{\pi \over 2},\pi } \right)$

$(\overrightarrow r - \overrightarrow c ) \times \overrightarrow b = 0$

$\overrightarrow r = \lambda \overrightarrow b + \overrightarrow c $

$ \Rightarrow \lambda \overrightarrow b \,.\,\overrightarrow a + \overrightarrow c \,.\,\overrightarrow a = 0$

$ \Rightarrow \lambda (3) + (7 + 8) = 0$

$ \Rightarrow \lambda = - 5$

$\overrightarrow r = 5\overrightarrow b + \overrightarrow c $

$ = - 5\widehat i - 5\widehat j - 5\widehat k + (7\widehat i + 3\widehat j + 4\widehat k)$

$ = 2\widehat i - 8\widehat j - \widehat k$

$\therefore$ $\overrightarrow r \,.\,\overrightarrow c = 17 + 24 - 4 = 34$

$[\matrix{ {\overrightarrow c } & {\overrightarrow a } & {\overrightarrow b } \cr } ] = - 25$

Let $\overrightarrow c = l\widehat i + n\widehat j + n\widehat k$

$\left| {\matrix{ l & m & n \cr 4 & 3 & 0 \cr 3 & { - 4} & 5 \cr } } \right| = - 25$

$ \Rightarrow 3l - 4m - 5n = - 5$ ..... (i)

$\overrightarrow c \,.\,(\widehat i + \widehat j + \widehat k) = 4$

$ \Rightarrow l + m + n = 4$ ..... (ii)

${{\overrightarrow c \,.\,\overrightarrow a } \over {|\overrightarrow a |}} = 1 \Rightarrow \overrightarrow c \,.\,\overrightarrow a = 5$

$ \Rightarrow 4l + 3m = 5$ ...... (iii)

Using (i), (ii) and (iii)

$l = 2,m = - 1,n = 3$

Now, ${{\overrightarrow c \,.\,\overrightarrow b } \over {|\overrightarrow b |}} = {{25} \over {5\sqrt 2 }} = {5 \over {\sqrt 2 }}$

$\therefore$ Option (2) is correct.

$\vec{a}=\lambda \hat{i}+\mu \hat{j}+4 \hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}-2 \hat{k}, \vec{c}=2 \hat{i}+3 \hat{j}+\hat{k}$

Now, $\vec{a} \cdot \vec{b}=\sqrt{54} \Rightarrow \frac{-2 \lambda+4 \mu-8}{\sqrt{24}}=\sqrt{54}$

$\Rightarrow-2 \lambda+4 \mu-8=36$

$\Rightarrow 2 \mu-\lambda=22\quad...(i)$

and $\left|\begin{array}{ccc}\lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1\end{array}\right|=0$

$10 \lambda-2 \mu-56=0 \quad...(ii)$

By (i) & (ii) $\lambda=\frac{78}{9}, \mu=\frac{138}{9}$

$\therefore \mu+\lambda=24$

$ \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}) $

Taking cross product with $\vec{a}$

$ \begin{aligned} & \Rightarrow \vec{a} \times(\vec{a} \times \vec{b})=\vec{a} \times(\hat{i}-\hat{j}) \\\\ & \Rightarrow (\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b}=\hat{i}+\hat{j}+2 \hat{k} \\\\ & \Rightarrow \vec{a}-3 \vec{b}=\hat{i}+\hat{j}+2 \hat{k} \\\\ & \Rightarrow 2 \vec{a}-6 \vec{b}=2 \hat{i}+2 \hat{j}+4 \hat{k} \\\\ & \Rightarrow \vec{a}-6 \vec{b}=3 \hat{i}+3 \hat{j}+3 \hat{k} \end{aligned} $

Let $\mathrm{A}:(3,-4,2) \quad \mathrm{C}:(-2,-1,3)$

$ \text { B : }(1,2,-1) \quad \text { D: }(5,-2 \alpha, 4) $

A, B, C, D are coplanar points, then

$ \begin{aligned} & \Rightarrow\left|\begin{array}{ccc} 1-3 & 2+4 & -1-2 \\ -2-3 & -1+4 & 3-2 \\ 5-3 & -2 \alpha+4 & 4-2 \end{array}\right|=0 \\\\ & \Rightarrow \alpha=\frac{73}{17} \end{aligned} $

First, we write $\overrightarrow{b}$ as a linear combination of $\overrightarrow{a}$ and $\overrightarrow{j}$ since $\overrightarrow{b}$ is a rotation of $\overrightarrow{a}$ about the y-axis.

$\vec{b}=\lambda \vec{a}+\mu \hat{j}=\lambda(-\hat{i}+2 \hat{j}+\hat{k})+\mu \hat{j}=-\lambda \hat{i}+(2 \lambda+\mu \hat{j})+\lambda \hat{k}$

$\overrightarrow{b}$ is orthogonal to $\overrightarrow{a}$ due to the right angle rotation, so $\overrightarrow{b} \cdot \overrightarrow{a} = 0$.

This implies :

$\begin{aligned} & (-\hat{i}+2 \hat{j}+\hat{k}) \cdot(-\lambda \hat{i}+(2 \lambda+\mu) \hat{j}+\lambda \hat{k})=0 \\\\ & \lambda+2(2 \lambda+\mu)+\lambda=0 \Rightarrow 6 \lambda+2 \mu=0 \Rightarrow \mu+3 \lambda=0\end{aligned}$

$ \therefore $ $\vec{b}=\lambda \vec{a}-3 \lambda \hat{j}=\lambda(-\hat{i}+2 \hat{j}+\hat{k})-3 \lambda \hat{j}=\lambda(-\hat{i}-\hat{j}+\hat{k})$

The magnitude of $\overrightarrow{b}$ is the same as the magnitude of $\overrightarrow{a}$ because a rotation doesn't change the magnitude of a vector. This gives us :

$ |\vec{b}|=\sqrt{3}|\lambda|=\sqrt{6}[\because|a|=\sqrt{6}] \Rightarrow|\lambda|=\sqrt{2} \Rightarrow \lambda \neq \sqrt{2} $

as for this value of $\lambda$ angle between $b$ and $y$-axis is not acute.

Therefore $\lambda=-\sqrt{2}$

Thus, we have :

$\overrightarrow{b} = -\sqrt{2}(-\hat{i} - \hat{j} + \hat{k}) = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} - \sqrt{2}\hat{k}$

Then, we find the vector $3\overrightarrow{a} + \sqrt{2}\overrightarrow{b}$ :

$3\overrightarrow{a} + \sqrt{2}\overrightarrow{b} = 3(-\hat{i} + 2\hat{j} + \hat{k}) + \sqrt{2}(\sqrt{2}\hat{i} + \sqrt{2}\hat{j} - \sqrt{2}\hat{k}) $

$= -3\hat{i} + 6\hat{j} + 3\hat{k} + 2\hat{i} + 2\hat{j} - 2\hat{k} = -\hat{i} + 8\hat{j} + \hat{k}$

The projection of this vector onto $\overrightarrow{c}$ is given by the dot product divided by the magnitude of $\overrightarrow{c}$ :

$\frac{(-\hat{i} + 8\hat{j} + \hat{k}) \cdot (5\hat{i} + 4\hat{j} + 3\hat{k})}{\sqrt{5^2 + 4^2 + 3^2}} = \frac{-5 + 32 + 3}{\sqrt{50}} = \frac{30}{5\sqrt{2}} = 3\sqrt{2}$

So the projection of $3\overrightarrow{a} + \sqrt{2}\overrightarrow{b}$ onto $\overrightarrow{c}$ is $3\sqrt{2}$ which is option D.

$\vec{b}(\vec{a} \cdot \vec{c})-\vec{c}(\vec{a} \cdot \vec{b})=\frac{\vec{b}-\vec{c}}{2}$ $\vec{a} \cdot \vec{c}=\frac{1}{2}, \quad \vec{a} \cdot \vec{b}=\frac{1}{2}$

$ \begin{aligned} (\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d}) & =(\vec{b} \cdot \vec{d})(\vec{a} \cdot \vec{c})-(\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c}) \\\\ & =(\vec{a} \cdot \vec{b})(\vec{a} \cdot \vec{c}) \\\\ & =\frac{1}{4} \end{aligned} $

Let $\vec{\beta}_1=\lambda \vec{\alpha}$

Now $\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1$

$ \begin{aligned} & =(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})-\lambda(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \\\\ & =(1-4 \lambda) \hat{\mathrm{i}}+(2-3 \lambda) \hat{\mathrm{j}}-(5 \lambda+4) \hat{\mathrm{k}} \\\\ & \vec{\beta}_2 \cdot \vec{\alpha}=0 \\\\ & \Rightarrow 4(1-4 \lambda)+3(2-3 \lambda)-5(5 \lambda+4)=0 \\\\ & \Rightarrow 4-16 \alpha+6-9 \lambda-25 \lambda-20=0 \\\\ & \Rightarrow 50 \lambda=-10 \\\\ & \Rightarrow \lambda=\frac{-1}{5} \\\\ & \vec{\beta}_2=\left(1+\frac{4}{5}\right) \hat{\mathrm{i}}+\left(2+\frac{3}{5}\right) \hat{\mathrm{j}}-(-1+4) \hat{\mathrm{k}} \\\\ & \vec{\beta}_2=\frac{9}{5} \hat{\mathrm{i}}+\frac{13}{5} \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\\\ & 5 \vec{\beta}_2=9 \hat{\mathrm{i}}+13 \hat{\mathrm{j}}-15 \hat{\mathrm{k}} \\\\ & 5 \vec{\beta}_2 \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=9+13-15=7 \end{aligned} $

Let the position vector of $P, Q, R$ be $\overrightarrow{0}, \overrightarrow{a}, \overrightarrow{b}$

$\Rightarrow$ Position vector of $A=\frac{2 \overrightarrow{a}+\overrightarrow{b}}{3}, $

Position vector of $B=\frac{2 \overrightarrow{b}}{3}$ and

Position vector of $C=\frac{\overrightarrow{a}}{3}$

$ \begin{aligned} & \therefore \overrightarrow{A B}=\frac{2 \vec{b}}{3}-\left(\frac{2 \vec{a}+\vec{b}}{3}\right)=\frac{\vec{b}}{3}-\frac{2 \vec{a}}{3}=\frac{\vec{b}-2 \vec{a}}{3} \\\\ & \overrightarrow{C A}=\frac{2 \vec{a}+\vec{b}}{3}-\frac{\vec{a}}{3}=\frac{\vec{a}+\vec{b}}{3} \\\\ & \text { Area of } \triangle P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|=\frac{1}{2}|\vec{a} \times \vec{b}| \\\\ & \text { Area of } \triangle A B C=\frac{1}{2}|\overrightarrow{C A} \times \overrightarrow{A B}| \\ & =\frac{1}{2}\left|\left(\frac{\vec{a}+\vec{b}}{3}\right) \times\left(\frac{\vec{b}-2 \vec{a}}{3}\right)\right|=\frac{1}{2}\left|\frac{\vec{a} \times \vec{b}}{3}\right| \\\\ & \therefore \frac{\text { Area }(\triangle P Q R)}{\text { Area }(\triangle A B C)}=\frac{\frac{1}{2}|\vec{a} \times \vec{b}|}{\frac{1}{2}\left|\frac{\vec{a} \times \vec{b}}{3}\right|}=3 \text { sq. units } \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \vec{v} \times \vec{w}=(\vec{u}+\lambda \vec{v})=\hat{i}-\hat{j}-2 \hat{k}+\lambda(2 \hat{i}+\hat{j}-\hat{k}) \\\\ & =(2 \lambda+1) \hat{i}+(\lambda-1) \hat{j}-(2+\lambda) \hat{k} \\ & \end{aligned}\\ &\begin{aligned} & \text { Now, } \vec{v} \cdot(\vec{v} \times \vec{w})=0 \\\\ & \Rightarrow(2 \hat{i}+\hat{j}-\hat{k}) \cdot((2 \lambda+1) \hat{i}+(\lambda-1) \hat{j}-(\lambda+2) \hat{k})=0 \\\\ & \Rightarrow2(2 \lambda+1)+\lambda-1+\lambda+2=0 \Rightarrow 6 \lambda+3=0 \\\\ & \Rightarrow \lambda=-\frac{1}{2} \\\\ & \vec{w} \cdot(\vec{v} \times \vec{w})=\vec{w} \cdot(\vec{u}+\lambda \vec{v})=\vec{u} \cdot \vec{w}+\lambda \vec{v} \cdot \vec{w}=0 \\\\ & \vec{u} \cdot \vec{w}=-\lambda \vec{v} \cdot \vec{w} \\\\ & \vec{u} \cdot \vec{w}=\frac{1}{2} \times 2=1 \end{aligned} \end{aligned} $

$|\vec{a}||\vec{b}||\vec{c}|=14$

$ \begin{aligned} & \vec{a} \wedge \vec{b}=\vec{b} \wedge \vec{c}=\vec{c} \wedge \vec{a}=\theta=\frac{2 \pi}{3} \\\\ & \vec{a} \cdot \vec{b}=-\frac{1}{2}|\vec{a}||\vec{b}| \\\\ & \vec{b} \cdot \vec{c}=-\frac{1}{2}|\vec{b}||\vec{c}| \\\\ & \vec{c} \cdot \vec{a}=-\frac{1}{2}|\vec{c}||\vec{a}| \end{aligned} $

Now,

$ \begin{aligned} & \begin{aligned} &(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b}) \\\\ &=168 \quad\quad...(i)\\\\ &(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c})-(\vec{a} \cdot \vec{c})|\vec{b}|^{2} \\\\ &= \frac{1}{4}|\vec{b}|^{2}|\vec{a}||\vec{c}|+\frac{1}{2}|\vec{a}||\vec{b}|^{2}|\vec{c}| \\\\ &= \frac{3}{4}|\vec{a}||\vec{b}|^{2}|\vec{c}| \quad\quad...(ii) \end{aligned} \end{aligned} $

Similarly $(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})=\frac{3}{4}|\vec{a}||\vec{b}||\vec{c}|^{2} \quad\quad...(iii)$

$(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b})=\frac{3}{4}|\vec{a}|^{2}|\vec{b}||\vec{c}| \quad\quad...(iv)$

Substitute (ii), (iii), (iv) in (i)

$\frac{3}{4}|\vec{a}||\vec{b}||\vec{c}|[|\vec{a}|+|\vec{b}|+|\vec{c}|]=168$

$\frac{3}{4} \times 14[|\vec{a}|+|\vec{b}|+|\vec{c}|]=168$

$|\vec{a}|+|\vec{b}|+|\vec{c}|=16$

$\overrightarrow a = 3\widehat i + \widehat j$ & $\overrightarrow b = \widehat i + 2\widehat j + \widehat k$

$\overrightarrow a \times (\overrightarrow b \times \overrightarrow c ) = (\overrightarrow a \,.\,\overrightarrow c )\overrightarrow b - (\overrightarrow a \,.\,\overrightarrow b )\overrightarrow c = \overrightarrow b + \lambda \overrightarrow c $

If $\overrightarrow b $ & $\overrightarrow c $ are non-parallel

then $\overrightarrow a \,.\,\overrightarrow c = 1$ & $\overrightarrow a \,.\,\overrightarrow b = - \lambda $

but $\overrightarrow a \,.\,\overrightarrow b = 5 \Rightarrow \lambda = - 5$

$\widehat a\,.\,\widehat b = {1 \over {\sqrt 2 }}$ and $|\widehat a \times \widehat b| = {1 \over {\sqrt 2 }}$

${{\left( {\widehat a + \widehat b} \right)\,.\,\left( {\widehat a + 2\widehat b + 2\left( {\widehat a \times \widehat b} \right)} \right)} \over {\left| {\widehat a + \widehat b} \right|\left| {\widehat a + 2\widehat b + 2\left( {\widehat a \times \widehat b} \right)} \right|}} = \cos \theta $

$ \Rightarrow \cos \theta = {{1 + 3\widehat a\widehat b + 2} \over {|\widehat a + \widehat b||\widehat a + 2\widehat b + 2(\widehat a \times \widehat b)|}}$

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

$|\widehat a + 2\widehat b + 2(\widehat a \times \widehat b){|^2} = 1 + 4 + 4|\widehat a \times \widehat b{|^2} + 4\widehat a\widehat b$

$ = 5 + 4\,.\,{1 \over 2} + {4 \over {\sqrt 2 }} = 7 + 2\sqrt 2 $

So, ${\cos ^2}\theta = {{{{\left( {3 + {3 \over {\sqrt 2 }}} \right)}^2}} \over {(2 + \sqrt 2 )(7 + 2\sqrt 2 )}} = {{9\sqrt 2 (5\sqrt 2 + 3)} \over {164}}$

$ \Rightarrow 164{\cos ^2}\theta = 90 + 27\sqrt 2 $

$\overrightarrow u = a({\log _e}b)\widehat i - 6\widehat j + 3\widehat k$

$\overrightarrow v = ({\log _e}b)\widehat i + 2\widehat j + 2a({\log _e}b)\widehat k$

For acute angle $\overrightarrow u \,.\,\overrightarrow v > 0$

$ \Rightarrow a{({\log _e}b)^2} - 12 + 6a({\log _e}b) > 0$

$\because$ $b > 1$

Let ${\log _e}b = t \Rightarrow t > 0$ as $b > 1$

$a{t^2} + 6at - 12 > 0\,\,\,\,\,\,\,\forall t > 0$

$ \Rightarrow a \in \phi $

Clearly $\overrightarrow a $, $\overrightarrow b $, $\overrightarrow c $ are non-coplanar

$\left| {\matrix{ {1 + t} & {1 - t} & 1 \cr {1 - t} & {1 + t} & 2 \cr t & { - t} & 1 \cr } } \right| \ne 0$

$ \Rightarrow (1 + t)(1 + t + 2t) - (1 - t)(1 - t - 2t) + 1({t^2} - t - t - {t^2}) \ne 0$

$ \Rightarrow (3{t^2} + 4t + 1) - (1 - t)(1 - 3t) - 2t \ne 0$

$ \Rightarrow (3{t^2} + 4t + 1) - (3{t^2} - 4t + 1) - 2t \ne 0$

$ \Rightarrow t \ne 0$

$\left( {x\overrightarrow a + y\overrightarrow b } \right).\left( {6y\overrightarrow a - 18x\overrightarrow b } \right) = 0$

$ \Rightarrow \left( {6xy|\overrightarrow a {|^2} - 18xy|\overrightarrow b {|^2}} \right) + \left( {6{y^2} - 18{x^2}} \right)\overrightarrow a .\overrightarrow b = 0$

As given equation is identity

Coefficient of ${x^2} = $ coefficient of ${y^2} = $ coefficient of $xy = 0$

$ \Rightarrow |\overrightarrow a {|^2} = 3|\overrightarrow b {|^2} \Rightarrow |\overrightarrow b | = 3\sqrt 3 $

and $\overrightarrow a .\overrightarrow b = 0$

$|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a ||\overrightarrow b |\sin \theta $

$ = 9.\,3\sqrt 3 .1 = 27\sqrt 3 $

$\overrightarrow a = \alpha \widehat i + \widehat j + \beta \widehat k$, $\overrightarrow b = 3\widehat i - 5\widehat j + 4\widehat k$

$\overrightarrow a \times \overrightarrow b = - \widehat i + 9\widehat j + 12\widehat k$

$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr \alpha & 1 & \beta \cr 3 & { - 5} & 4 \cr } } \right| = - \widehat i + 9\widehat j + 12\widehat k$

$4 + 5\beta = - 1 \Rightarrow \beta = - 1$

$ - 5\alpha - 3 = 12 \Rightarrow \alpha = - 3$

$\overrightarrow b - 2\overrightarrow a = 3\widehat i - 5\widehat j + 4\widehat k - 2\left( { - 3\widehat i + \widehat j - \widehat k} \right)$

$\overrightarrow b - 2\overrightarrow a = 9\widehat i - 7\widehat j + 6\widehat k$

$\overrightarrow b + \overrightarrow a = \left( {3\widehat i - 5\widehat j + 4\widehat k} \right) + \left( { - 3\widehat i + \widehat j - \widehat k} \right)$

$\overrightarrow b + \overrightarrow a = - 4\widehat j + 3\widehat k$

Projection of $\overrightarrow b - 2\overrightarrow a $ on $\overrightarrow b + \overrightarrow a $ is $ = {{\left( {\overrightarrow b - 2\overrightarrow a } \right)\,.\,\left( {\overrightarrow b + \overrightarrow a } \right)} \over {\left| {\overrightarrow b + \overrightarrow a } \right|}}$

$ = {{28 + 18} \over 5} = {{46} \over 5}$

Given, $\overrightarrow a = 2\widehat i - \widehat j + 5\widehat k$ and $\overrightarrow b = \alpha \widehat i + \beta \widehat j + 2\widehat k$

Also, $\left( {\left( {\overrightarrow a \times \overrightarrow b } \right) \times i} \right)\,.\,\widehat k = {{23} \over 2}$

$ \Rightarrow \left( {\left( {\overrightarrow a \,.\,\widehat i} \right)\overrightarrow b - \left( {\overrightarrow b \,.\,\widehat i} \right)\,.\,\overline a } \right)\,.\,\widehat k = {{23} \over 2}$

$ \Rightarrow \left( {2\,.\,\overrightarrow b - \alpha \,.\,\overrightarrow a } \right)\,.\,\widehat k = {{23} \over 2}$

$ \Rightarrow 2\,.\,2 - 5\alpha = {{23} \over 2} \Rightarrow \alpha = {{ - 3} \over 2}$

Now, $\left| {\overrightarrow b \times 2j} \right| = \left| {\left( {\alpha \widehat i + \beta \widehat j + 2\widehat k} \right) \times 2\widehat j} \right|$

$ = \left| {2\alpha \widehat k + 0 - 4\widehat i} \right|$

$ = \sqrt {4{\alpha ^2} + 16} $

$ = \sqrt {4{{\left( {{{ - 3} \over 2}} \right)}^2} + 16} = 5$

Given : $\overrightarrow a = (\alpha ,1, - 1)$ and $\overrightarrow b = (2,1, - \alpha )$

$\overrightarrow c = \overrightarrow a \times \overrightarrow b = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr \alpha & 1 & { - 1} \cr 2 & 1 & { - \alpha } \cr } } \right|$

$ = ( - \alpha + 1)\widehat i + ({\alpha ^2} - 2)\widehat j + (\alpha - 2)\widehat k$

Projection of $\overrightarrow c $ on $\overrightarrow d = - \widehat i + 2\widehat j - 2\widehat k$

$ = \left| {\overrightarrow c \,.\,{{\overrightarrow d } \over {|d|}}} \right| = 30$ {Given}

$ \Rightarrow \, = \left| {{{\alpha - 1 - 4 + 2{\alpha ^2} - 2\alpha + 4} \over {\sqrt {1 + 4 + 4} }}} \right| = 30$

On solving $\alpha = {{ - 13} \over 2}$ (Rejected as $\alpha > 0$)

and $\alpha = 7$

$\overrightarrow a = \widehat i - \widehat j + 2\widehat k$

$\overrightarrow a \times \overrightarrow b = 2\widehat i - \widehat k$

$\overrightarrow a \,.\,\overrightarrow b = 3$

$|\overrightarrow a \times \overrightarrow b {|^2} + |\overrightarrow a \,.\,\overrightarrow b {|^2} = |\overrightarrow a {|^2}\,.\,|\overrightarrow b {|^2}$

$ \Rightarrow 5 + 9 = 6|\overrightarrow b {|^2}$

$ \Rightarrow |b {|^2} = {7 \over 3}$

$|\overrightarrow a - \overrightarrow b | = \sqrt {|\overrightarrow a {|^2} + |\overrightarrow b {|^2} - 2\overrightarrow a \,.\,\overrightarrow b } = \sqrt {{7 \over 3}} $

projection of $\overrightarrow b $ on $\overrightarrow a - \overrightarrow b = {{\overrightarrow b \,.\,(\overrightarrow a - \overrightarrow b )} \over {|\overrightarrow a - \overrightarrow b |}}$

$ = {{\overrightarrow b \,.\,\overrightarrow a - |\overrightarrow b {|^2}} \over {|\overrightarrow a - \overrightarrow b |}} = {{3 - {7 \over 3}} \over {\sqrt {{7 \over 3}} }}$

$ = {2 \over {\sqrt {21} }}$

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

then $\bar{a}+\vec{c}=-\vec{b}$

then $(\vec{a}+\vec{c}) \times \vec{b}=-\vec{b} \times \bar{b}$

$\therefore \quad \vec{a} \times \vec{b}+\vec{c} \times \vec{b}=\overline{0}\quad\dots(i)$

For $(S 1):|\vec{a} \times \vec{b}+\vec{c} \times \vec{b}|-|\vec{c}|=6(2 \sqrt{2}-1)$

$ \begin{aligned} &|(\vec{a}+\vec{c}) \times \vec{b}|-|\vec{c}|=6(2 \sqrt{2}-1) \\\\ &|\vec{c}|=6-12 \sqrt{2} \text { (not possible) } \end{aligned} $

Hence (S1) is not correct

For (S2) : from (i) $\vec{b}+\vec{c}=-\vec{a}$

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

$\Rightarrow 12+12=-6 \sqrt{2} \cdot 2 \sqrt{3} \cos (\pi-\angle A C B)$

$\therefore \cos (\angle A C B)=\sqrt{\frac{2}{3}}$

$ \begin{aligned} &\therefore \angle A C B=\cos ^{-1} \sqrt{\frac{2}{3}} \\\\ &\therefore S(2) \text { is correct. } \end{aligned} $

Given,

$\overrightarrow a = - \widehat i + \widehat j + \widehat k$

$\overrightarrow b = 2\widehat i + \widehat j - \widehat k$

and let $\overrightarrow c = x\widehat i + y\widehat j + z\widehat k$

Now, $\overrightarrow a + \overrightarrow b = \widehat i + 2\widehat j$

and $\overrightarrow a \times \overrightarrow b = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1} & 1 & 1 \cr 2 & 1 & { - 1} \cr } } \right| = - 2\widehat i + \widehat j - 3\widehat k$

$\overrightarrow c $ is coplanar with $\overrightarrow a $ and $\overrightarrow b $

$\therefore$ $\left[ {\overrightarrow c \,\overrightarrow a \,\overrightarrow b } \right] = 0$

$ \Rightarrow \left| {\matrix{ x & y & z \cr { - 1} & 1 & 1 \cr 2 & 1 & { - 1} \cr } } \right| = 0$

$ \Rightarrow x( - 2) - y( - 1) + z( - 3) = 0$

$ \Rightarrow - 2x + y - 3z = 0$ ..... (1)

Now, $\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow a \times \overrightarrow b } \right)$

$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 2 & 0 \cr { - 2} & 1 & { - 3} \cr } } \right|$

$ = - 6\widehat i + 3\widehat j + 5\widehat k$

Given, $\overrightarrow c \,.\,\left[ {\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right] = - 42$

$ \Rightarrow \left( {x\widehat i + y\widehat j + z\widehat k} \right)\,.\,\left( { - 6\widehat i + 3\widehat j + 5\widehat k} \right) = - 42$

$ \Rightarrow - 6x + 3y + 5z = - 42$ ...... (2)

Now, $\overrightarrow a - \overrightarrow b = \left( { - \widehat i + \widehat j + \widehat k} \right) - \left( {2\widehat i + \widehat j - \widehat k} \right)$

$ = - 3\widehat i + 0\widehat j + 2\widehat k$

$\therefore$ $\overrightarrow c \times \left( {\overrightarrow a - \overrightarrow b } \right)$

$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr x & y & z \cr { - 3} & 0 & 2 \cr } } \right|$

$ = 2y\widehat i - \widehat j(2x + 3z) + 3y\widehat k$

Given,

$\left( {\overrightarrow c \times \left( {\overrightarrow a - \overrightarrow b } \right)} \right)\,.\,\widehat k = 3$

$ \Rightarrow \left( {2y\widehat i - \widehat j(2x + 3z) + 3y\widehat k} \right)\,.\,\widehat k = 3$

$ \Rightarrow 3y = 3$

$ \Rightarrow y = 1$

Putting value of $y = 1$ in equation (1) and (2) we get,

$ - 2x + 1 - 3z = 0$ ..... (3)

and $ - 6x + 3 + 5z = - 42$

$ \Rightarrow - 6x + 5z = - 45$ ..... (4)

Solving (3) and (4), we get

$x = 5$ and $z = - 3$

$\therefore$ $\overrightarrow c = 5\widehat i + \widehat j - 3\widehat k$

$ \Rightarrow {\left| {\overrightarrow c } \right|^2} = {5^2} + {1^2} + {( - 3)^2} = 35$

$\overrightarrow{A B} \| \overrightarrow{A C}$ if

$\frac{1}{2}=\frac{\alpha-4}{-6}=\frac{1}{2}$

$\Rightarrow \alpha=1$

$\vec{a}, \vec{b}, \vec{c}$ are non-collinear for $\alpha=2$ (smallest positive integer)

Mid point of $B C=M\left(\frac{5}{2}, 0, \frac{9}{2}\right)$

$A M=\sqrt{\frac{9}{4}+16+\frac{9}{4}}=\frac{\sqrt{82}}{2}$

$\overrightarrow a = \alpha \widehat i + 3\widehat j - \widehat k$

$\overrightarrow b = 3\widehat i - \beta \widehat j + 4\widehat k$

$\overrightarrow c = \widehat i + 2\widehat j - 2\widehat k$

Projection of $\overrightarrow a $ on $\overrightarrow c $ is

${{\overrightarrow a \,.\,\overrightarrow c } \over {|\overrightarrow b |}} = {{10} \over 3}$

${{\alpha + 6 + 2} \over {\sqrt {{1^2} + {2^2} + {{( - 2)}^2}} }} = {{\alpha + 8} \over 3} = {{10} \over 3}$

$\therefore$ $\alpha$ = 2

$\overrightarrow b \times \overrightarrow c = - 6\widehat i + 10\widehat j + 7\widehat k$

$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & { - \beta } & 4 \cr 1 & 2 & { - 2} \cr } } \right| = (2\beta - 8)\widehat i + 10\widehat j + (6 + \beta )\widehat k = - 6\widehat i + 10\widehat j + 7\widehat k$

$2\beta - 8 = - 6$ & $6 + \beta = 7$

$\therefore$ $\beta$ = 1

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

$\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$ and $\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$

$\therefore$ $\overrightarrow a \times \overrightarrow b = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr \alpha & 2 & { - 1} \cr { - 2} & \alpha & 1 \cr } } \right| = (2 + \alpha )\widehat i - (\alpha - 2)\widehat j + ({\alpha ^2} + 4)\widehat k$

Now $\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {15({\alpha ^2} + 4)} $

$ \Rightarrow {(2 + \alpha )^2} + {(\alpha - 2)^2} + {({\alpha ^2} + 4)^2} = 15({\alpha ^2} + 4)$

$ \Rightarrow {\alpha ^4} - 5{\alpha ^2} - 36 = 0$

$\therefore$ $\alpha = \, \pm \,3$

Now, $2{\left| {\overrightarrow a } \right|^2} + \left( {\overrightarrow a - \overrightarrow b } \right){\left| {\overrightarrow b } \right|^{ - 2}} = 2.14 - 14 = 14$

Let $\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$

and $\overrightarrow a \,.\,\left( {3\widehat i - {1 \over 2}\widehat j + 2\widehat k} \right) = 0 \Rightarrow 3{a_1} + {{{a_2}} \over 2} + 2{a_3} = 0$ ..... (i)

and $\overrightarrow a \times (2\widehat i + \widehat k) = 2\widehat i - 13\widehat j - 4\widehat k$

$ \Rightarrow {a_2}\widehat i + (2{a_3} - {a_1})\widehat j - 2{a_2}\widehat k = 2\widehat i - 13\widehat j - 4\widehat k$

$\therefore$ ${a_2} = 2$ ..... (ii)

and ${a_1} - 2{a_3} = 13$ ..... (iii)

From eq. (i) and (iii) : ${a_1} = 3$ and ${a_3} = - 5$

$\therefore$ $\overrightarrow a = 3\widehat i + 2\widehat j - 5\widehat k$

$\therefore$ projection of $\overrightarrow a $ on $2\widehat i + 2\widehat j + \widehat k = {{6 + 4 - 5} \over 3} = {5 \over 3}$

$\because$ $\overrightarrow a $ and $\overrightarrow b $ be the vectors along the diagonals of a parallelogram having area 2$\sqrt2$.

$\therefore$ ${1 \over 2}|\overrightarrow a \times \overrightarrow b | = 2\sqrt 2 $

$|\overrightarrow a ||\overrightarrow b |\sin \theta = 4\sqrt 2 $

$ \Rightarrow |\overrightarrow b |\sin \theta = 4\sqrt 2 $ ..... (i)

and $|\overrightarrow a \,.\,\overrightarrow b | = |\overrightarrow a \times \overrightarrow b |$

$|\overrightarrow a ||\overrightarrow b |\cos \theta = |\overrightarrow a ||\overrightarrow b |\sin \theta $

$ \Rightarrow \tan \theta = 1$

$\therefore$ $\theta = {\pi \over 4}$

By (i) $|\overrightarrow b | = 8$

Now $\overrightarrow c = 2\sqrt 2 (\overrightarrow a \times \overrightarrow b ) - 2\overrightarrow b $

$ \Rightarrow \overrightarrow c \,.\,\overrightarrow b = - 2|\overrightarrow b {|^2} = - 128$ ...... (ii)

and $\overrightarrow c \,.\,\overrightarrow c = 8|\overrightarrow a \times \overrightarrow b {|^2} + 4|\overrightarrow b {|^2}$

$ \Rightarrow |\overrightarrow c {|^2} = 8.32 + 4.64$

$ \Rightarrow |\overrightarrow c | = 16\sqrt 2 $ ..... (iii)

From (ii) and (iii)

$|\overrightarrow c ||\overrightarrow b |\cos \alpha = - 128$

$ \Rightarrow \cos \alpha = {{ - 1} \over {\sqrt 2 }}$

$\alpha = {{3\pi } \over 4}$