Vector Algebra

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

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

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

Let $c=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}$

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

$\begin{aligned} & \therefore [\mathbf{a} \mathbf{b} \mathbf{c}]=0 \\ & \Rightarrow \quad \left|\begin{array}{ccc} 2 & 1 & 1 \\ 1 & 2 & -1 \\ x & y & z \end{array}\right|=0 \\ & \Rightarrow x(-1-2)-y(-2-1)+z(4-1)=0 \end{aligned}$

$\Rightarrow \quad x-y-z=0$ .... (i)

$\because \mathrm{c}$ is perpendicular to a.

$\because \quad \mathbf{a} \cdot \mathbf{c}=0 \Rightarrow 2 x+y+z=0$ .... (ii)

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

$\begin{aligned} & \frac{x}{0}=\frac{y}{3}=\frac{z}{-3}=\lambda \\ & \Rightarrow x=0, y=3 \lambda \text { and } z=-3 \lambda \\ & \text { But, }|c|=1 \Rightarrow x^2+y^2+z^2=1 \\ & \Rightarrow \quad 9 \lambda^2+9 \lambda^2=1 \Rightarrow \lambda= \pm \frac{1}{3 \sqrt{2}} \\ & \therefore \quad c= \pm \frac{1}{\sqrt{2}}(-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \end{aligned}$

Thus, we have $\mathbf{c}=\frac{1}{\sqrt{2}}(-\hat{\mathbf{j}}+\hat{k})$

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

$\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} }}$

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$

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

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

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

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 }}$

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

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

$\overrightarrow c = 2\widehat i - 3\widehat j + 2\widehat k$

Now, $\overrightarrow b \times \overrightarrow c = \overrightarrow a $

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

$\overrightarrow c \,.\,\overrightarrow a = 0$

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

$ = 2 - 3 - 2 = 0$

$ \Rightarrow - 3 = 0$ (Not possible)

$\Rightarrow$ No possible value of $\overrightarrow b $ is possible.

$\because$ $\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = 3\overrightarrow b - \overrightarrow c = \overrightarrow u $

$\overrightarrow b \times \left( {\overrightarrow c \times \overrightarrow a } \right) = \overrightarrow c - 2\overrightarrow a = \overrightarrow v $

$\overrightarrow c \times \left( {\overrightarrow b \times \overrightarrow a } \right) = 3\overrightarrow b - 2\overrightarrow a = \overrightarrow w $

$\therefore$ $\overrightarrow u + \overrightarrow v = \overrightarrow w $

So, vectors $\overrightarrow u $, $\overrightarrow v $ and $\overrightarrow w $ are coplanar, hence their Scalar triple product will be zero.

${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1 \Rightarrow {\cos ^2}\alpha = {1 \over 3} \Rightarrow \cos \alpha = {1 \over {\sqrt 3 }}$

$\overrightarrow a = {\lambda \over 3}(\widehat i + \widehat j + \widehat k),\,\lambda > 0$

${\lambda \over {\sqrt 3 }}{{(\widehat i + \widehat j + \widehat k)\,.\,(3\widehat i + 4\widehat j)} \over {\sqrt {{3^2} + {4^2}} }} = 7$

$ \Rightarrow {\lambda \over {\sqrt 3 }}(3 + 4) = 7 \times 5$

$\therefore$ $\lambda = 5\sqrt 3 $

$\overrightarrow a = 5(\widehat i + \widehat j + \widehat k)$

Let $\overrightarrow b = p\widehat i + q\widehat j + r\widehat k$

$\overrightarrow a \,.\,\overrightarrow b = 0$ and $[\overrightarrow a \,\overrightarrow b \,\widehat i] = 0$

$ \Rightarrow p + q + r = 0$ ..... (i)

& $\left| {\matrix{ p & q & r \cr 1 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right| = 0 \Rightarrow \matrix{ {q = r} \cr {p = - 2r} \cr } $

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

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

Now $\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right|$

$5\sqrt 3 = \left| r \right|\sqrt b \Rightarrow \left| r \right| = {5 \over {\sqrt 2 }}$

$\Rightarrow$ Projection of $\overrightarrow b $ on $3\widehat i + 4\widehat j = \left| {{{\overrightarrow b \,.\,\left( {3\widehat i + 4\widehat j} \right)} \over {\sqrt {{3^2} + {4^2}} }}} \right|$

$ = \left| r \right|{{( - 6 + 4)} \over 5} = \left| {{{ - 2r} \over 5}} \right|$

Projection $ = {2 \over 5} \times {5 \over {\sqrt 2 }} = \sqrt 2 $

$\therefore$ B is correct.

$\left| {\widehat a + \widehat b + 2(\widehat a \times \widehat b)} \right| = 2,\,\theta \in (0,\,\pi )$

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

$ \Rightarrow {\left| {\widehat a} \right|^2} + {\left| {\widehat b} \right|^2} + 4{\left| {\widehat a \times \widehat b} \right|^2} + 2\widehat a\,.\,\widehat b = 4$

$\therefore$ $\cos \theta = \cos 2\theta $

$\therefore$ $\theta = {{2\pi } \over 3}$

where $\theta$ is angle between $\widehat a$ and $\widehat b$.

$\therefore$ $2\left| {\widehat a \times \widehat b} \right| = \sqrt 3 = \left| {\widehat a - \widehat b} \right|$

(S1) is correct.

And projection of $\widehat a$ on $(\widehat a + \widehat b) = \left| {{{\widehat a\,.\,(\widehat a + \widehat b)} \over {\left| {\widehat a + \widehat b} \right|}}} \right| = {1 \over 2}$

(S2) is correct.

Given,

$\left[ {\matrix{ 0 & { - {c_3}} & {{c_2}} \cr {{c_3}} & 0 & { - {c_1}} \cr { - {c_2}} & {{c_1}} & 0 \cr } } \right]\left[ {\matrix{ 1 \cr {{b_2}} \cr {{b_3}} \cr } } \right] = \left[ {\matrix{ {3 - {c_1}} \cr {1 - {c_2}} \cr { - 1 - {c_3}} \cr } } \right]$

$\therefore$ $ - {b_2}{c_3} + {b_3}{c_2} = 3 - {c_1}$ ..... (1)

${c_3} - {b_3}{c_1} = 1 - {c_2}$ ...... (2)

$ - {c_2} + {b_2}{c_1} = - 1 - {c_3}$ ..... (3)

Adding (1), (2) and (3), we get

$({b_2} - {b_3}){c_1} + ({b_3} - 1){c_2} + (1 - {b_2}){c_3}$

$ = 3 - ({c_1} + {c_2} + {c_3})$

$ \Rightarrow ({c_2}{b_3} - {c_3}{b_2}) - ({c_1}{b_3} - {c_3}) + ({c_1}{b_2} - {c_2})$

$ = 3 - ({c_1} + {c_2} + {c_3})$ ..... (4)

Also given,

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

$\overrightarrow b = \widehat i + {b_2}\widehat j + {b_3}\widehat k$

$\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k$

Now,

$\overrightarrow c \times \overrightarrow b $

$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr {{c_1}} & {{c_2}} & {{c_3}} \cr 1 & {{b_2}} & {{b_3}} \cr } } \right|$

$ = ({c_2}{b_3} - {c_3}{b_2})\widehat i - \widehat j({c_1}{b_3} - {c_3}) + \widehat k({c_1}{b_2} - {c_2})$

And

$\overrightarrow a - \overrightarrow c = (3 - {c_1})\widehat i + (1 - {c_2})\widehat j + ( - 1 - {c_3})\widehat k$

Comparing value of $\overrightarrow c \times \overrightarrow b $ and $\overrightarrow a - \overrightarrow c $ with equation (4), we get

$\overrightarrow c \times \overrightarrow b = \overrightarrow a - \overrightarrow c $ ..... (5)

Multiplying both side with $\overrightarrow b $, we get

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

$ \Rightarrow 0 = 0 - \overrightarrow b \,.\,\overrightarrow c $ [Given $\overrightarrow a \,.\,\overrightarrow b = 0$]

$ \Rightarrow \overrightarrow b \,.\,\overrightarrow c = 0$

$\therefore$ (B) is correct.

Now multiplying both side of equation (5) with $\overrightarrow c $, we get

$\overrightarrow c \,.\,(\overrightarrow c \times \overrightarrow b ) = \overrightarrow a \,.\,\overrightarrow c - |\overrightarrow c {|^2}$

$ \Rightarrow 0 = \overrightarrow a \,.\,\overrightarrow c - |\overrightarrow c {|^2}$

$ \Rightarrow \overrightarrow a \,.\,\overrightarrow c = |\overrightarrow c {|^2}$

Now given $\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k$

$\therefore$ $|\overrightarrow c {|^2} = \sqrt {c_1^2 + c_2^2 + c_3^2} $

$ \Rightarrow |\overrightarrow c {|^2} = c_1^2 + c_2^2 + c_3^2$

$\therefore$ $|\overrightarrow c {|^2} = 0$ when ${c_1} = 0$, ${c_2} = 0$ and ${c_3} = 0$

Now if we put ${c_1} = {c_2} = {c_3} = 0$ in matrix we get,

$\left[ {\matrix{ 0 & 0 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right]\left[ {\matrix{ 1 \cr {{b_2}} \cr {{b_3}} \cr } } \right] = \left[ {\matrix{ {3 - {c_1}} \cr {1 - {c_2}} \cr { - 1 - {c_3}} \cr } } \right]$

$ \Rightarrow \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right] = \left[ {\matrix{ 3 \cr 1 \cr { - 1} \cr } } \right]$

$\therefore$ Matrix does not satisfy the value of ${c_1} = {c_2} = {c_3} = 0$.

$\therefore$ ${c_1},{c_2},{c_3}$ can't be zero.

$\therefore$ $|\overrightarrow c {|^2} \ne 0$

$ \Rightarrow \overrightarrow a \,.\,\overrightarrow c \ne 0$

$\therefore$ Option (A) is wrong.

As $\overrightarrow a \,.\,\overrightarrow c = |\overrightarrow c {|^2}$

$ \Rightarrow |\overrightarrow a ||\overrightarrow c |\cos \theta = |\overrightarrow c {|^2}$ [$\theta$ = angle between $\overrightarrow a $ and $\overrightarrow c $]

$ \Rightarrow |\overrightarrow a |\cos \theta = |\overrightarrow c |$

$ \Rightarrow (\sqrt {9 + 1 + 1} )\cos \theta = |\overrightarrow c |$

$ \Rightarrow |\overrightarrow c | = \sqrt {11} \cos \theta $

$ \Rightarrow |\overrightarrow c | \le \sqrt {11} $ [as $ - 1 \le \cos \theta \le 1$]

$\therefore$ Option (D) is correct.

Given, $\overrightarrow a \,.\,\overrightarrow b = 0$

$ \Rightarrow (3\widehat i + \widehat j - \widehat k)\,.\,(\widehat i + {b_2}\widehat j + {b_3}\widehat k) = 0$

$ \Rightarrow 3 + {b_2} - {b_3} = 0$

$ \Rightarrow {b_3} = 3 + {b_2}$

Now,

$\overrightarrow b = \widehat i + {b_2}\widehat j + {b_3}\widehat k$

$ \Rightarrow |\overrightarrow b | = \sqrt {1 + b_2^2 + b_3^2} $

$ \Rightarrow |\overrightarrow b {|^2} = 1 + b_2^2 + {(3 + {b_2})^2}$

$ \Rightarrow |\overrightarrow b {|^2} = 1 + b_2^2 + 9 + 6{b_2} + b_2^2$

$ \Rightarrow |\overrightarrow b {|^2} = 10 + 6{b_2} + 2b_2^2$

$ \Rightarrow |\overrightarrow b {|^2} = 10 + 2{b_2}(3 + {b_2})$

$ \Rightarrow |\overrightarrow b {|^2} = 10 + 2{b_2}{b_3}$

$ \Rightarrow |\overrightarrow b {|^2} > 10$ [as ${b_2}{b_3} > 0$ then $2{b_2}{b_3} > 0$]

$ \Rightarrow |\overrightarrow b | > \sqrt {10} $

$\therefore$ Option (C) is correct.

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

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

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

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

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

On comparing corresponding components,

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

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

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

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

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

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

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

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

We know,

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

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

According to the question,

Using section formula,

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

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

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

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

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

Median through vertex $A$ is

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

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

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

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

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

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

Statement I is false.

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

Statement II is also false.

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

∴ Position vector of $\mathbf{C}$ is

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Let A be taken as origin.

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

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

Position vector of

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Let $\theta$ be the required angle.

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

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

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

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

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

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

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

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

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

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

Hence, three vectors are possible.