Mathematics M.C.Q (1 Marks)

$\because {{a}}.\,{{b}} = 0,\,\,\,\,\,\,\therefore {{a}}\, \bot \,{{b}}$

So, angle between ${{a}}$ and ${{b}}$ is $\theta = 90^o$

Similarly, $[{{a}}\,{{b}}\,{{c}}] = |{{a}}||{{b}}|{{\hat n}}.\,{{c}}$, where ${{\hat n}}$ is a normal vector

 $\therefore [{{a}}\,{{b}}\,{{c}}] = |{{a}}||{{b}}|{{\hat n}}\,\,\,{{c}}$

$\because {{\hat n}}$  and ${{c}}$ are parallel to each other

 $\therefore [{{a}}\,{{b}}\,{{c}}] = |{{a}}||{{b}}||{{\hat n}}|\,\,\,\,|{{c}}|\cos \theta $$ = |{{a}}||{{b}}||{{c}}|$.

= $a(3 \times 4\sin \frac{{2\pi }}{3}.\hat n)$= $a.(12 \times \frac{{\sqrt 3 }}{2}\hat n)$

= $6\sqrt 3 |a||\hat n|$= $6\sqrt 3 \times 2 \times 1 \Rightarrow 12\sqrt 3 $.

==> $\alpha ,\,\beta ,\,\gamma $ are coplanar

==> $[\alpha \,\,\beta \,\,\gamma ] = 0$.
 

Therefore, $(r - {a_1}).({a_1} \times {a_2}) = 0$

==> $[r\,\,{a_1}\,{a_2}]$=$[{a_1}\,{a_1}\,{a_2}] \Rightarrow [r\,{a_1}\,{a_2}] = 0$

Hence, the required plane is $[r\,\,{a_1}\,{a_2}] = 0$.

$r.(a \times b + b \times c + c \times a) = [a\;b\;c]$

Therefore, the length of the perpendicular from the origin to this plane is given by

$= \frac{{[a\,b\,c]}}{{|a \times b + b \times c + c \times a|}}$.

Hence, its equation is $(r - a).((b - a) \times c) = 0$
or

$r.(b \times c + c \times a) = [a\,b\,c\,]$

The length of the perpendicular from the origin to this plane is

$=\frac{{[a\,b\,c]}}{{|b \times c + c \times a|}}$.

$ = \frac{1}{6}|(2i - 2j) \times (2i - 2k)\,.\,(i - j - k)|$

$ = \frac{1}{6}\left| {\,\begin{array}{*{20}{c}}2&{ - 2}&0\\2&0&{ - 2}\\1&{ - 1}&{ - 1}\end{array}\,} \right| = \frac{1}{6}( - 4) = - \frac{2}{3} = \frac{2}{3}$ cubic unit.

$\therefore $ $x\,.\,(y \times z) = 0$ $ \Rightarrow $ $[x\,\,y\,\,z] = 0$

==> $\left| {\,\begin{array}{*{20}{c}}2&1&\alpha \\\alpha &0&1\\5&{ - 1}&0\end{array}\,} \right| = 0 \Rightarrow \alpha = \pm \,\sqrt 7 $.

$\left.\quad+\left(2 \lambda-\lambda^{2}-3 \lambda\right) \hat{k}\right]=0$

$\Rightarrow \quad 2-2 \lambda-3+\lambda+2 \lambda-\lambda^{2}-3 \lambda=0$

$\Rightarrow \quad-\lambda^{2}-2 \lambda-1=0$

$\Rightarrow \quad(\lambda+1)^{2}=0$

$\Rightarrow \quad \lambda=-1$

New volume $=[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}]$

$=\left[\begin{array}{lll}{\vec{a}} & {\vec{b}} & {\vec{c}}\end{array}\right]^{2}=4$

$\Rightarrow\left|\begin{array}{ccc}{-1} & {-1} & {0} \\ {-\lambda-2} & {0} & {0} \\ {1} & {-2} & {-2}\end{array}\right|=0 \Rightarrow \lambda=-2$

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

$=1-\hat{j} \cdot \hat{j}+1$

$=1-1+1$

$=1$

The correct answer is $A.$

$i \times (a \times i) + j \times (a \times j) + k \times (a \times k)$

$ = (i.i)a - i(a.i) + (j.j)a - j(a.j) + (k.k)a - k(a.k)$

$ = 3a - a = 2a.$

$ = \left| {\begin{array}{*{20}{c}}i&j&k\\1&2&{ - 2}\\{ - 2}&3&7\end{array}} \right| = 20i - 3j + 7k.$

$ = {(\alpha)^2}(\gamma\,.\,\beta) - (\gamma\,.\,\alpha)(\beta\,.\,\alpha)$$ = 14( - 3) - (4)(8) = - 74$.

Let $a \times b = d$

so , $(b \times c)[(c \times a) \times d] = (b \times c)[(d.\,a)c - (d.c).a]$

$ = (b \times c)[a.(a \times b).c - (a \times b)c.a]$

$ = (b \times c)[a\,b\,c]a = a.[b \times c].[a\,b\,c]$

$ = [a\,b\,c][a\,b\,c] = {[a\,b\,c]^2}$.

$\because \,\,a\, \bot \,\,b\,\,,\,\,\,\therefore \,\,\,a\,.\,b = 0$

$\because a|\,|c$, $\therefore \,\,\,a\,.\,c = 1$ $(a, b$  and  $ c $ are unit vectors)

$a \times (b \times c) = (1)b - (0)\,c\,$=$ b.$

$ = a \times [a \times {a^2}b] = a \times {a^3}b\,\hat n = |a{|^4}b.$

$ = (a\,.\,b)(a \times a) - (a\,.\,a)(a \times b)$$ = (a\,.\,b)\,0 + (a\,.\,a)(b \times a)$

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

Now, $b = r \times a = \left\{ {xa + yb + z(a \times b)} \right\} \times a$

$ = y(b \times a) + z[(a \times b) \times a]$$ = - y(a \times b) - z\,[a \times (a \times b)]$

$ = - y(a \times b) - z\,[(a\,.\,b)a - (a\,.\,a)\,b]$

$ = - y(a \times b) + z(a\,.\,a)b$,

$ \Rightarrow y = 0$ and $z = \frac{1}{{(a\,.\,a)}} \Rightarrow r = xa + \frac{1}{{a\,.\,a}}(a \times b)$.

$ \Rightarrow 0 = b\,.\,a + \lambda \,|a{|^2} \Rightarrow \lambda = - \frac{{b\,.\,\,a}}{{|a{|^2}}} = \frac{5}{6}$

Also, $(a \times r) \times a = b \times a + \lambda \,a \times a \Rightarrow r = \frac{7}{6}i + \frac{2}{3}j$.

$\therefore \,\,\,a + b + c + d = (\alpha + 1)d$ and $a + b + c + d = (\beta + 1)\,a.$

$ \Rightarrow (\alpha + 1)\,d = (\beta + 1)\,a$

If $\alpha \, \ne \, - 1,$ then $(\alpha + 1)\,d = (\beta + 1)\,a \Rightarrow d = \frac{{\beta + 1}}{{\alpha + 1}}a$

$ \Rightarrow a + b + c = \alpha \,d \Rightarrow a + b + c = \alpha \left( {\frac{{\beta + 1}}{{\alpha + 1}}} \right)a$

$ \Rightarrow \left( {1 - \frac{{\alpha (\beta + 1)}}{{\alpha + 1}}} \right)\,a + b + c = 0$

$ \Rightarrow a,\,b,\,c$ are coplanar which is contradiction to the given condition, $\therefore \,\,\,\alpha = - 1$ and so $a + b + c + d = 0.$

Squaring both sides, we get

${a^2}{l^2} + {m^2}{b^2} + {n^2}{c^2} + 2l\,m\,a\,.\,b + 2l\,n\,a\,.\,c + 2m\,n\,b\,.\,c = 0$

But $a,\,b,\,c$ are mutually perpendicular

So, $a\,.\,b,$ $b\,.\,c$ and $c\,.\,a$ are equal to zero.
Therefore, ${a^2}{l^2} + {m^2}{b^2} + {n^2}{c^2} = 0\,\,i.e.,\,\,l,\,\,m,\,\,n$are equal to zero because ${a^2},\,\,{b^2}$ and ${c^2}$ cannot be equal to zero.

$ = \frac{{{a^2} + {b^2} - 2a\,.\,b}}{{{a^2}{b^2}}} = {\left( {\frac{{a - b}}{{ab}}} \right)^2}.$

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

$=-3 \hat{i}-4 \hat{j}-3 \hat{j}-3 \hat{k}=-3 \hat{i}-7 \hat{j}-3 \hat{k}$

Now $\overrightarrow{ v } \times(3 \hat{ i }+7 \hat{ j }+3 \hat{ k })=\overrightarrow{0}$

$\Rightarrow \overrightarrow{ v }=\lambda(3 \hat{ i }+7 \hat{ j }+3 \hat{ k })$

Also $\overrightarrow{ v } \cdot \overrightarrow{ j }=-7 \Rightarrow 7 \lambda=-7 \Rightarrow \lambda=-1$

$\Rightarrow \overrightarrow{ v }=-3 \hat{ i }-7 \hat{ j }-3 \hat{ k }$

so, $\overrightarrow{ v } \cdot \hat{i}=-3$

Since, $| r - b |+| r - c |=4$ is equation of ellipsoid in plane $r \cdot a =5$ or $x+y+z=5$.

$\therefore$ Distance between foci $=| b - c |$

$\text { and } 2 a=4 \quad=|3 \hat{ i }-\hat{ j }-2 \hat{ k }|=\sqrt{14}=2 a e$

$\therefore \quad e=\frac{\sqrt{14}}{4} \Rightarrow e^2=1-\frac{b^2}{a^2}$

$\Rightarrow \frac{14}{16}=1-\frac{b^2}{4} \Rightarrow \frac{b^2}{4}=\frac{2}{16}$

$b^2=\frac{1}{2}$

$\therefore$ Area of ellipsoid

$=\pi a b=\pi(2)\left(\frac{1}{\sqrt{2}}\right)=\sqrt{2} \pi$

And from the given options the closest integer to $\sqrt{2} \pi$ is $4$.

Given, $\left(\sum_{i=1}^3 \frac{ a _i}{2^i}\right)^2=\sum_{i=1}^3 \frac{a_i^2}{2^i}$

$\Rightarrow\left(\frac{a_1}{2}+\frac{a_2}{4}+\frac{a_3}{8}\right)^2=\frac{a_1^2}{2}+\frac{a_2^2}{4}+\frac{\iota_3^2}{8}$

$\Rightarrow \frac{a_1^2}{4}+\frac{a_2^2}{16}+\frac{ a _3^2}{64}+\frac{ c _1 a_2}{4}+\frac{a_2 u _3}{16}+\frac{ a _3 a_1}{8}$

$\Rightarrow \frac{a_1^2}{4}+\frac{3 a_2^2}{16}+\frac{7 a_3^2}{64}=\frac{a_1 a_2}{4}+\frac{a_2 a_3}{16}+\frac{a_3 a_1}{8}$

$\Rightarrow 16 a_1^2+12 a_2^2+7 a_3^2$

$=16 a_1 a_2+4 a_2 a_3+8 a_3 a_1$

$\Rightarrow\left(8 a_1^2+8 a_2^2-16 a_1 a_2\right)+\left(4 a_2^2+a_3^2-4 a_2 a_3\right)$

$+\left(2 a_3^2+8 a_1^2-8 a_3 a_1\right)+4 a_3^2=0$

$\Rightarrow 8\left(a_1-a_2\right)^2+\left(2 a_2-a_3\right)^2+2\left(a_3-2 a_1\right)^2$ $+4 a_3^2=0$

The above relation will be true if $a_1=a_2, 2 a_2=a_3, a_3=2 a$ and $a_3=0$

So, $a_1=a_2=a_3=0$

Then, set $A$ contains exactly one element.

Let position vector of $\triangle A B C$ are $A(\vec{a}), B(\vec{b})$ and $C(\vec{c})$

Let circumeentre of $\triangle A B C, O$ (origin).

Centroid of $\triangle A B C$ is $\frac{\vec{a}+\vec{b}+\vec{c}}{3}$

We know that centroid divide orthocentre and circumcentre in $2: 1$

$H G: G O=2: 1$

$\overrightarrow{O G}=\left(\frac{\overrightarrow{ a }+\vec{b}+\vec{c}}{3}\right)$

$\overrightarrow{O H}=\vec{a}+\vec{b}+\vec{c}$

$\vec{N}$ is mid-point of $\overrightarrow{O H}$

$\vec{N}=\frac{\vec{a}+\vec{b}+\vec{c}}{2}$

$\overrightarrow{N A}+\overrightarrow{N B}+\overrightarrow{N C}$

$=\vec{a}-\left(\frac{\vec{a}+\vec{b}+\vec{c}}{2}\right)+\vec{b}-\left(\frac{\vec{a}+\vec{b}+\vec{c}}{2}\right)$ $+\vec{c}-\left(\frac{\vec{a}+\vec{b}+\vec{c}}{2}\right)$

$=(\vec{a}+\vec{b}+\vec{c})-\frac{3(\vec{a}+\vec{b}+\vec{c})}{2}$

$=-\frac{1}{2}(\vec{a}+\vec{b}+\vec{c})=-\frac{1}{2} \overrightarrow{O H}=\frac{1}{2} \overrightarrow{H O}$

We have,

$| v - i |=| v -2 i |=| v - j |$

Clearly, $v$ is circumcentre of $\triangle A B C$.

where $A(1,0), B(2,0), C(0,1)$ and $O(x, y)$

$\therefore \quad O A^2=O B^2=O C^2$

$(x-1)^2+(y-0)^2=(x-2)^2+y^2$

$=x^2+(y-1)^2$

$\Rightarrow x^2-2 x+1+y^2=x^2-4 x+4+y^2$

$=x^2+y^2-2 y+1$

$\Rightarrow \quad 2 x=3 \Rightarrow x=\frac{3}{2}$

$\Rightarrow x^2-2 x+1+y^2=x^2+y^2-2 y+1$

$\Rightarrow \quad x=y=3 / 2$

$\therefore \quad(x, y)=\left(\frac{3}{2}, \frac{3}{2}\right)$

$\Rightarrow \quad \vec{v}=\frac{3}{2} \hat{i}+\frac{3}{2} \hat{j}$

$|\vec{v}|=\sqrt{\frac{9}{4}+\frac{9}{4}}=\frac{3 \sqrt{2}}{2}$ lie in $(2,3]$

Given,

$\vec{a}=6 \hat{i}-3 \hat{j}-6 \hat{k}, \vec{d}=\hat{i}+\hat{j}+\hat{k}$ and $b$ is parallel to $d$ and cis perpendicular to $d$.

$\therefore \quad b=\lambda d$ and $\vec{c} \cdot \vec{d}=0$

$b=\lambda(\hat{i}+\hat{j}+\hat{k})$

Now, $\quad \overrightarrow{ a }=\vec{b}+\vec{c}$

$\Rightarrow \vec{c}=\vec{e}-\vec{b}=(6 \hat{i}-3 \hat{j}-6 \hat{k})-\lambda(\hat{i}+\hat{j}+\hat{k})$

$\Rightarrow c=(6-\lambda) \hat{i}-(3+\lambda) \hat{j}-(6+\lambda) \hat{k}$

$\Rightarrow \quad \vec{c} \cdot \vec{d}=0$

$\therefore \quad((6-\lambda) \hat{i}-(3+\lambda) \hat{j}-(6+\lambda) \hat{k})$

$\therefore 6-\lambda-3-\lambda-6=\lambda=0-j \lambda+k)=0$

$\rightarrow 6-\lambda-3-\lambda-6-\lambda=0 \Rightarrow \lambda=-1$

$\therefore \quad c=7 \hat{i}-2 \hat{j}-5 \hat{k}$

We have, $v _1, v _2, v _3, v _4$ be unit vectors lie in $X Y$-plane.

Let

$v _1=\cos \alpha i +\sin \alpha j \alpha \in\left(0,90^{\circ}\right)$

$v _2=\cos \beta i +\sin \beta j \beta \in\left(90^{\circ}, 180^{\circ}\right)$

$v _3=\cos \gamma i +\sin \gamma j \gamma \in\left(180^{\circ}, 270^{\circ}\right)$

$v _1=\cos \delta i +\sin \delta j \delta \in\left(270^{\circ}, 360^{\circ}\right)$

(a) $v _1+ v _2+ v _3+ v _4=0$ not necessarily true for all $v _1, v _2, v _3$ and $v _4$

(b) $v _i+ v _j(1 \leq i < j \leq 4)$

$v _1+ v _2=(\cos \alpha+\cos \beta) i +(\sin \alpha+\sin \beta) j$

If $y$ coordinate is positive i.e.

$\alpha$ in 1st quadrant $\beta$ in $2$ nd quadrant.

But in this case $x$-coordinate is not necessarily positive.

(c) $\quad v _i= v _j(1 \leq i < j \leq 4)$

Let $\alpha$ in 1st quadrant, $\gamma$ in 3 rd quadrant $\therefore \quad v _i \cdot v _j<0$

(d) $v _i, v _j=(\cos \alpha \cos \beta)+\sin \alpha \sin \beta$

$=\cos (\alpha-\beta)$ is positive $0<|\alpha-\beta|<\frac{\pi}{2}$

We have, The position vector of the vertices of $\triangle A B C, A(\vec{a}), B(\vec{b})$ and $C(\vec{c})$ respectively.

Given,

$\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a} \neq 0$

$\vec{a} \times \vec{b}=\vec{b} \times \vec{c}$

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

$(\vec{a}+\vec{c}) \times \vec{b}=0$

$\vec{e}+\vec{c}=\lambda \vec{b}$

${[\because \overrightarrow{\vec{e}}+\vec{c}}$ is parallel to ${\vec{b}] }$

We have,

$\vec{u}=2 \hat{i}-\hat{j}+\hat{k} \text { and } \vec{v}=-3 \hat{j}+2 \hat{k}$

$\vec{u} \times \vec{v}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 0 & -3 & 2\end{array}\right|$

$=(-2+3) \hat{i}-(4-0) \hat{j}+(-6-0) \hat{k}$

$\vec{u} \times \vec{v}=\hat{i}-4 \hat{j}-6 \hat{k}$

$\vec{w}$ is a unit vector in $X Y$-plane.

Let $\vec{w}=\cos \theta \hat{i}+\sin \theta \hat{j}$

$\therefore|(\vec{u} \times \vec{v}) \cdot \vec{w}|=\mid(\hat{i}-4 \hat{j}-6 \hat{k})$ $=\cos \theta-4 \sin \theta$

$\therefore \text { Maximum value of }$

$\cos \theta-4 \sin \theta=\sqrt{1+4^2}=\sqrt{17}$

Given,

Point $P$ inside the $\triangle A B C$ such that

$P A +2 P B +3 P C =0$

Let $\quad P A = a - P$

$PB = b - P$

PC $=$ c - P

$\therefore( a - P )+2( b - P )+3( c - P )=0$

$\Rightarrow \quad a +2 b +3 c =3 P$

$\Rightarrow \quad P =\frac{ a +2 b +3 c }{6}$

Area of $\triangle A P C=\frac{1}{2} \mid a \times P + P \times c+c \times$ a

$=\frac{1}{2}\left| a \times \frac{( a +2 b +3 c )}{6}+\frac{( a +2 b +3 c )}{6}\right|$

$=\frac{1}{2 \times 6} \mid 2(a \times b)+3(a \times c)+(a \times c)+2$

$( b \times c )+6 c \times a \mid$

$=\frac{1}{6}| a \times b + b \times c + c \times a |$

$\frac{\text { Area of } \triangle A B C}{\text { Area of } \triangle A P C}=\frac{\frac{1}{2}|a \times b+b \times c+c \times a|}{\frac{1}{6}|a \times b+b \times c+c \times a|}$

$=3$

$\therefore$ Ratio $=3: 1$

We have,

$a =3 i -4 k$ and $b =5 j +12 k$

We know that,

angle bisector of vector $a$ and $b$

$=\lambda\left[\frac{ a }{| a |}+\frac{ b }{| b |}\right]$

$\therefore \lambda\left[\frac{3 \hat{ i }-4 \hat{ k }}{5}+\frac{5 \hat{ j }+12 \hat{ k }}{13}\right]$

$=\lambda\left[\frac{39 \hat{ i }-52 \hat{ k }+25 \hat{ j }+60 \hat{ k }}{65}\right]$

$=\lambda[39 \hat{ i }+2 \hat{ j}+8 \hat{ k }]$

$AC=5$

$\therefore \mathrm{D}$ is midpoint of $\mathrm{BC}$

$ \mathrm{D}\left(\frac{1}{2}, \frac{3}{2}, 3\right) $

$ \therefore l=\sqrt{\left(2-\frac{1}{2}\right)^2+\left(-3-\frac{3}{2}\right)^2+(3-3)^2} $

$ l=\sqrt{\frac{45}{2}} $

$ \therefore 2 l^2=45$

$ \overrightarrow{\mathrm{b}}+6 \overrightarrow{\mathrm{c}}=\mu \overrightarrow{\mathrm{a}}$

Eliminating $\vec{a}$

$ \lambda \overrightarrow{\mathrm{c}}-5 \overrightarrow{\mathrm{b}}=\frac{6}{\mu} \overrightarrow{\mathrm{c}}+\frac{1}{\mu} \overrightarrow{\mathrm{b}} $

$ \therefore \mu=\frac{-1}{5}, \lambda=-30 $

$ \alpha=5, \beta=30$

$ |\vec{a}|^2-|\vec{b}|^2=0 \rightarrow 1+\lambda^2+9=9+1+4 $

$ \therefore \lambda=2, \cos \theta=\frac{\vec{a}-\vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\frac{3-\lambda-6}{\sqrt{14} \cdot \sqrt{14}} $

$ 14 \cos \theta=3-8=-5 $

$ \therefore(14 \cos \theta)^2=25$

$ \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=3 $

$ \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=\mathrm{k}(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}) $

$ 3=\mathrm{k}(2+6-15+3-2+3 \lambda) $

$ 3=\mathrm{k}(-6+3 \lambda) $ ...............($1$)

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

$ |\overrightarrow{\mathrm{r}}|=\mathrm{k} \sqrt{25+4+25+\lambda^2-10 \lambda}=1 . $        ...............($2$)

$ \mathrm{k}=\frac{3}{-6+3 \lambda}=\frac{1}{-2+\lambda} \quad \text { put in }(2) $

$ 4+\lambda^2-4 \lambda=54+\lambda^2-10 \lambda $

$ 6 \lambda=50 $

$ 3 \lambda=25$

$=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 5 & -1 & 7 \\ 1 & 2 & 3\end{array}\right|$

$=\frac{1}{2}|-17 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}|=\frac{1}{2} \sqrt{474}$

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

$|(\vec{b} \times \vec{a})-\vec{b}|^2=|\vec{b} \times \vec{a}|^2+|\vec{b}|^2 $

$=4+1=5$

$ \mathrm{B}(5,7,-6) $

$ \mathrm{C}(3,4,-10) $

$ \mathrm{D}(-1,-4,-2) $

 Area $=\frac{1}{2}|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}|=\frac{1}{2}|(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}) \times(6 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})| $

$ =\frac{1}{2}|112 \hat{\mathrm{i}}-64 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}| $

$ =4|14 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}-\hat{\mathrm{k}}| $

$ =4 \sqrt{196+64+1} $

$ =4 \sqrt{261} $

$ =12 \sqrt{29}$

Area $=\frac{1}{2}|\overline{\mathrm{BD}} \times \overline{\mathrm{AC}}|$

$ \overline{\mathrm{BD}}=\frac{5}{3} \hat{\mathrm{i}}-\frac{5}{3} \hat{\mathrm{j}}-\frac{2}{3} \hat{\mathrm{k}} $

$ \overline{\mathrm{AC}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$

Area of parallelogram having sides

$ \overrightarrow{\mathrm{OA}} \& \overrightarrow{\mathrm{OC}}=|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OC}}|=|2 \overrightarrow{\mathrm{a}} \times 3 \overrightarrow{\mathrm{b}}|=15 $

$ 6|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=15 $

$ \Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\frac{5}{2} \ldots \ldots .(1)$

Area of quadrilateral

$ \mathrm{OABC}=\frac{1}{2}\left|\overrightarrow{\mathrm{d}}_1 \times \overrightarrow{\mathrm{d}}_2\right| $

$ =\frac{1}{2}|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{OB}}|=\frac{1}{2}|(3 \overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}}) \times(6 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{b}})| $

$ =\frac{1}{2}|18 \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}-10 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=14|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| $

$ =14 \times \frac{5}{2}=35$

$ \cos \theta=\frac{\alpha^2-4 \alpha-4}{\sqrt{\alpha^2+8} \sqrt{5 \alpha^2+4}} $

$ \Rightarrow \alpha^2-4 \alpha-4>0 \quad \Rightarrow(\alpha-2)^2>8 $

$ \Rightarrow \alpha^2-4 \alpha+4>8 \quad \alpha-2<-2 \sqrt{2} $

$ \Rightarrow \alpha-2>2 \sqrt{2} \text { or } \alpha-2 \sqrt{2} $

$ \alpha>2+2 \sqrt{2} \text { or } \alpha<2-2 \sqrt{2} $

$ \alpha \in(-\infty,-0.82) \cup(4.82, \infty)$

Least positive integral value of $\alpha \Rightarrow 5$

$ \overrightarrow{\mathrm{p}}_1=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}, \overrightarrow{\mathrm{p}}_2=\frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}$

$ \overrightarrow{\mathrm{p}}_3=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}}$

Now angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_1=\frac{\pi}{2}$ $\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_1=0 \Rightarrow \frac{\mathrm{x}}{\sqrt{2}}+\frac{\mathrm{z}}{\sqrt{2}}=0$

$\Rightarrow \mathrm{x}+\mathrm{z}=0$ $.........(i)$

Angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_2=\frac{\pi}{3}$

$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_2=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_2\right| \cos \frac{\pi}{3}$

$ \Rightarrow \frac{\mathrm{y}}{\sqrt{2}}+\frac{\mathrm{z}}{\sqrt{2}}=\frac{1}{2} \Rightarrow \mathrm{y}+\mathrm{z}=\frac{1}{\sqrt{2}}$   $..............(ii)$

Angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_3=\frac{2 \pi}{3}$

$ \hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_3=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_3\right| \cos \frac{2 \pi}{3} $

$ \Rightarrow \frac{\mathrm{x}}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{-1}{2} \Rightarrow \mathrm{x}+\mathrm{y}=\frac{-1}{\sqrt{2}}$    $............(iii)$

from equation $(i)$, $(ii)$ and $(iii)$ we get

$x=\frac{-1}{\sqrt{2}} \quad y=0 \quad z=\frac{1}{\sqrt{2}}$

Thus $\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}-\frac{1}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}$

$\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-2}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}} $

$ \therefore|\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}|^2=\left(\sqrt{\frac{4}{2}+\frac{1}{2}}\right)^2=\frac{5}{2}$

$ \overrightarrow{\mathrm{AC}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} $

$ \mathrm{AB}=\sqrt{9+25+4}=\sqrt{38} $

$ \mathrm{AC}=\sqrt{4+9+25}=\sqrt{38} $

$ \overrightarrow{\mathrm{AD}}=-\frac{1}{2} \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\frac{3}{2} \hat{\mathrm{k}}=-\frac{1}{2}(\hat{\mathrm{i}}+8 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$

Length of projection of $\overrightarrow{\mathrm{AD}}$ on $\overrightarrow{\mathrm{AC}}$

$=\left|\frac{\overrightarrow{\mathrm{AD}} \cdot \overrightarrow{\mathrm{AC}}}{|\overrightarrow{\mathrm{AC}}|}\right|=\frac{37}{2 \sqrt{38}}$