Mathematics M.C.Q (1 Marks)

$\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}$      .....$(i)$

Given two lines

$\frac{x}{1}=\frac{y}{-1}=\frac{z}{1}$         .........$(ii)$

and $\frac{x-1}{0}=\frac{y+1}{0}=\frac{z}{1}$        .....$(iii)$

since the line $(i)$ is perpendicular to both the lines $(ii)$ and $(iii)$, therefore

$a-b+c=0$       ....$(iv)$

$-2 b+c=0$           .....$(v)$

From $(iv)$ and $(v)$ $c=2 b$ and $a+b=0$ which are not satisfy by options $(c)$ and $(d)$. Hence options $(c)$ and $(d)$ are rejected.

Thus point $\left(x_{1}, y_{1}, z_{1}\right)$ on the required line will be either $(0,0,0)$ or $(1,-1,0)$.

Now foot of the perpendicular from point $(0,0,0)$ to the line $(iii)$

$=(1,-2 r-1, r)$

The direction ratios of the line joining the points $(0,0,0)$ and $(1,-2 r-1, r)$ are  $1,-2 r-1, r$

Since sum of the $x$ and $y$ -coordinate of direction ratio of the required line is $0 .$

$\therefore 1-2 r-1-0, \Rightarrow r=0$

Hence direction ratio are $1,-1,0$

But the $z$-direction ratio of the required line is twice the $y$ -direction ratio of the required line

i.e. $0=2(-1),$ which is not true.

Hence the shortest line does not pass through the point $(0,0,0) .$ Therefore option $(a)$ is also rejected.

$\left| {\begin{array}{*{20}{c}}
{2 - 1}&{3 - 4}&{4 - 5}\\
1&1&{ - k}\\
k&2&1
\end{array}} \right| = 0$

$\left| {\begin{array}{*{20}{c}}
1&{ - 1}&{ - 1}\\
1&1&{ - k}\\
k&2&1
\end{array}} \right| = 0$

$\Rightarrow 1(1+2 k)+\left(1+k^{2}\right)-(2-k)=0$

$\Rightarrow k^{2}+2 k+k=0$

$\Rightarrow k^{2}+3 k=0$

$\Rightarrow \mathrm{k}=\mathrm{0},-3$

and $l_{2}+m_{2}+n_{2}=0$

and $l_{1}^{2}+m_{1}^{2}-n_{1}^{2}=0$ and

$l_{2}^{2}+m_{2}^{2}-n_{2}^{2}=0$

$\left(\because l+m+n=0 \text { and } l^{2}+m^{2}-n^{2}=0\right)$

Angle between lines, $\theta$ is $\cos \theta=l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}$

As given $l^{2}+m^{2}=n^{2}$ and $l+m=-n$ $\Rightarrow(-n)^{2}-2 l m=n^{2} \Rightarrow 2 l m=0$ or $l m=0$

So $l_{1} m_{1}=0, l_{2} m_{2}=0$

If $l_{1}=0, m_{1} \neq 0$ then $l_{1} m_{2}=0$

If $m_{1}=0, l_{1} \neq 0$ then $l_{2} m_{1}=0$

If $I_{2}=0, m_{2} \neq 0$ then $l_{2} m_{1}=0$

If $m_{2}=0, l_{2} \neq 0$ then $l_{1} m_{2}=0$

Also $l_{1} l_{2}=0$ and $m_{1} m_{2}=0$

$l^{2}+m^{2}-n^{2}=l^{2}+m^{2}+n^{2}-2 n^{2}=0$

$\Rightarrow 1-2 n^{2}=0 \Rightarrow n=\pm \frac{1}{\sqrt{2}}$

$\therefore n_{1}=\pm \frac{1}{\sqrt{2}}, n_{2}=\pm \frac{1}{\sqrt{2}}$

$\therefore \cos \theta=\frac{1}{2} \theta=60^{\circ}$ (acute angle)

$x=\sqrt{\lambda} y+(\sqrt{\lambda}-1) \Rightarrow y=\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}$          .....$(i)$

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

From $(i)$ and $(ii)$

$\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}=\frac{y-0}{1}=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1}$         ....$(A)$

The equation $(\mathrm{A})$ is the equation of line $\mathrm{L}_{1}$

Similarly equation ofline $\mathrm{L}_{2}$ is

$x-\frac{(1-\sqrt{\mu})}{\sqrt{\mu}}=\frac{y-0}{1}=\frac{z-\sqrt{\mu}}{1-\sqrt{\mu}}$     ....$(B)$

Since $\mathrm{L}_{1} \perp \mathrm{L}_{2},$ therefore

$\sqrt{\lambda} \sqrt{\mu}+1 \times 1+(\sqrt{\lambda}-1)(1-\sqrt{\mu})=0$

$\Rightarrow \sqrt{\lambda}+\sqrt{\mu}=0 \Rightarrow \sqrt{\lambda}=-\sqrt{\mu}$

$\Rightarrow \lambda=\mu$

$L _2: \frac{ x +16}{3}=\frac{ y +11}{2}=\frac{ z +4}{\gamma}= b$

$x = a -11=3 b -16 \quad \Rightarrow a -3 b =-5$     $. . . (1)$

$y =2 a -21=2 b -11 \Rightarrow 2 a -2 b =10$      $. . . (2)$

$z=3 a-29=b r-4 \Rightarrow 3 a-b \gamma=25$   $. . . (3)$

from $(1)$ and $(2) $

$a =10, b =5$

Now from $(3)$

$3(10)-5 \gamma=25 \quad \therefore \gamma=1$

$R _1 \equiv(-1,-1,1)$

$OR _1=-\hat{ i }-\hat{ j }+\hat{ k }$

$\overrightarrow{ n }=\left|\begin{array}{lll} i & j & k \\ 1 & 2 & 3 \\ 3 & 2 & 1\end{array}\right|=-4 \hat{ i }-(-8) \hat{ j }-4 \hat{ k }$

$\overrightarrow{ n }=-4 \hat{ i }+8 \hat{ j }+4 \hat{ k }=-4(\hat{ i }-2 \hat{ j }+\hat{ k })$

$\hat{ n }= \pm \frac{4(\hat{ i }-2 \hat{ j }+\hat{ k })}{4 \sqrt{6}}= \pm \frac{(\hat{ i }-2 \hat{ j }+\hat{ k })}{\sqrt{6}}$

$\overrightarrow{ OR } \cdot \hat{ n }= \pm(-\hat{ i }-\hat{ j }+\hat{ k })\left(\frac{\hat{ i }-2 \hat{ j }+\hat{ k }}{\sqrt{6}}\right)= \pm \frac{2}{\sqrt{6}}= \pm \sqrt{\frac{4}{6}}= \pm \sqrt{\frac{2}{3}}$

DR'S of $\mathrm{AF}=-1,1,1$

DR'S of $C E=1,1,-1$

DR'S of $\mathrm{BD}=1,-1,1$

Equation of $\mathrm{OG} \Rightarrow \frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}}{1}$

Equation of $A B \Rightarrow \frac{x-1}{1}=\frac{y}{-1}=\frac{z}{0}$

Normal to both the line's

$=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 0\end{array}\right|=\hat{i}+\hat{j}-2 \hat{k}$

$\overrightarrow{\mathrm{OA}}=\hat{i}$

$\text { S.D. }=\frac{|\hat{i} \cdot(\hat{i}+\hat{j}-2 \hat{k})|}{|\hat{i}+\hat{j}-2 \hat{k}|}=\frac{1}{\sqrt{6}}$

$L_2: \vec{r}_2=\hat{j}-\hat{k}+\mu(\hat{i}+\hat{k})$

Let system of planes are

$a x+b y+c z=0$     $. . . . . . . (1)$

$\because \text { It contain } L_1$

$\therefore a+b+c=0$    $. . . . . . (2)$

$\text { For largest possi }$

$\therefore a+c=0$    $. . . . . .(3)$

$\Rightarrow b=0$

For largest possible distance between plane ($1$) and $\mathrm{L}_2$ the line $\mathrm{L}_2$ must be parallel to plane ($1$)

$\therefore a+c=0$

$\Rightarrow b=0$

$\therefore$ Plane $\mathrm{H}_0: \mathrm{x}-\mathrm{z}=0$

Now $\mathrm{d}\left(\mathrm{H}_0\right)=\perp$ distance from point $(0,1,-1)$ on $\mathrm{L}_2$ to plane.

$\Rightarrow \mathrm{d}\left(\mathrm{H}_0\right)=\left|\frac{0+1}{\sqrt{2}}\right|=\frac{1}{\sqrt{2}}$

$\therefore \mathrm{P} \rightarrow 5$

$\text { for ' } Q \text { ' distance }=\left|\frac{2}{\sqrt{2}}\right|=\sqrt{2}$

$\therefore Q \rightarrow 4$

$\therefore(0,0,0)$ lies on plane

$\therefore R \rightarrow 3$

for 'S' $x=z ; y=z ; x=1$

$\therefore$ point of intersection $\mathrm{p}(1,1,1)$.

$\therefore \mathrm{OP}=\sqrt{1+1+1}=\sqrt{3}$

$\therefore \mathrm{S} \rightarrow 2$

$\therefore$ option $[B]$ is correct

So, $\overline{ AB }=(2 \mu+\lambda-1) \hat{ i }+(-\mu-2 \lambda) \hat{ j }+(2 \mu-2 \lambda) \hat{ k }$

and vector along their shortest distance $=2 \hat{i}+2 \hat{j}-\hat{k}$.

Hence, $\frac{2 \mu+\lambda-1}{2}=\frac{-\mu-2 \lambda}{2}=\frac{2 \mu-2 \lambda}{-1}$

$\Rightarrow \lambda=\frac{1}{9} \& \mu=\frac{2}{9}$

Hence, $A \equiv\left(\frac{8}{9}, \frac{2}{9}, \frac{2}{9}\right)$ and $B \equiv\left(\frac{4}{9},-\frac{2}{9}, \frac{4}{9}\right)$

$\Rightarrow$ Mid point of $AB \equiv\left(\frac{2}{3}, 0, \frac{1}{3}\right)$

Since $P , Q , R$ are collinear, $\overline{ PQ }$ iscollinear with $\overline{ QR }$

Hence $\frac{-\lambda}{1}=\frac{\mu}{1-\mu}=\frac{1}{v-1}$

For every $\mu \in R-\{0,1\}$ there exist unique $\lambda, v \in R$

Hence $Q$ cannot have coordinates $(0,1,1)$ and $(0,0,1)$.

$\frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{1}=\alpha $$\quad..............(1)$

$Q(\alpha, \alpha, 1)$

Direction ratio of $PQ$ are

$\lambda-\alpha, \lambda-\alpha, \lambda-1$

Since $PQ$ is perpendicular to $(1)$

$\therefore \quad \lambda-\alpha+\lambda-\alpha+0=0 $

$ \lambda=\alpha$

$\therefore \quad$ Direction ratio of $P Q$ are

$0,0, \lambda-1$

Another line is

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

$\therefore \quad R(-\beta, \beta,-1) $

$\therefore \quad \text { Direction ratio of PR are } $

$ \lambda+\beta, \lambda-\beta, \lambda+1$

Since $P Q$ is perpendicular to $(ii)$

$\therefore -\lambda-\beta+\lambda-\beta=0 $

$ \beta=0 $

$\therefore \quad R(0,0,-1) $

$\text { and } \text { Direction ratio of } P Q \text { are } \lambda, \lambda, \lambda+1 $

$\text { Since } P Q \perp PR $

$\therefore 0+0+\lambda^2-1=0 \Rightarrow \lambda= \pm 1 \Rightarrow B, C$

For $\lambda=1$ the point is on the line so it will be rejected.

$\Rightarrow \quad \lambda=-1$

$\ell: \frac{ x -0}{ a }=\frac{ y -0}{ b }=\frac{ z -0}{ c }= k$

This line $\ell$ is perpendicular to given line $\ell_1$ and $\ell_2$.

Hence

$a+2 b+2 c=0 $

$2 a+2 b+c=0 $

$\frac{a}{-2}=\frac{b}{3}=\frac{c}{-2}$

Hence equation of $\ell$ is $\frac{ x }{-2}=\frac{ y }{3}=\frac{ z }{-2}= k _1, \quad k _2$

for $\ell_1$ for $\ell_2$

Now $A \left(-2 k _1, 3 k _1,-2 k _1\right) \quad B \left(-2 k _2, 3 k _2,-2 k _2\right)$

Point A satisfied $\ell_1$

$-2 k_1 \hat{i}+3 k_1 \hat{j}-2 k_1 \hat{k}=(3+t) \hat{i}+(-1+2 t) \hat{j}+(4+2 t) \hat{k} $

$3+t=-2 k_1 \quad \ldots \ldots . .(1) $

$-1+2 t=3 k_1 \quad \ldots \ldots .(2) $

$4+2 t=-2 k_1 \quad \ldots \ldots .(3)$

$(2)$ and $(3)$ $-5=5 k_1 \Rightarrow k_1=-1 \Rightarrow A(2,-3,2)$

Let any point on $\ell_2(3+2 S, 3+2 S, 2+S)$

Given$\quad\sqrt{(1+2 S)^2+(6+2 S)^2+(S)^2}=\sqrt{17} $

$9 S^2+28 S+37=17 $

$9 S^2+28 S+20=0 $

$9 S^2+18 S+10 S+20=0 $

$9 S(S+2)+10(S+2)=0 $

$S=-2,-10 / 9$

Hence $(-1,-1,0),(7 / 9,7 / 9,8 / 9)$

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

$\left|\begin{array}{ccc}5-\alpha & 0 & 0 \\ 0 & 3-\alpha & -2 \\ 0 & -1 & 2-\alpha\end{array}\right|=0$

$(5-\alpha)((3-\alpha)(2-\alpha)-2)=0 $

$\left(\alpha^2-5 \alpha+6-2\right)=0 $

$(\alpha-5)\left(\alpha^2-5 \alpha+4\right)=0 $

$\alpha=1,4,5$

$\vec{a} \equiv(1,-1,0), \vec{c}=(-1,-1,0) $

$\vec{b}=2 \hat{i}+k \hat{j}+2 \hat{k} \quad \vec{d}=5 \hat{i}+2 \hat{j}+k \hat{k}$

Now $\left|\begin{array}{lll}2 & 0 & 0 \\ 2 & k & 2 \\ 5 & 2 & k\end{array}\right|=0 \quad \Rightarrow \quad k= \pm 2$

$\overrightarrow{ n }_1=\overrightarrow{ b }_1 \times \overrightarrow{ d }_1=6 \hat{ j }-6 \hat{ k } \text { for } k =2 $

$\overrightarrow{ n }_2=\overrightarrow{ b }_2 \times \overrightarrow{ d }_2=14 \hat{ j }+14 \hat{ k } \text { for } k =-2$

so the equation of planes are $(\vec{r}-\vec{a}) \vec{n}_1=0 \Rightarrow y-z=-1$ $\qquad$............$(1)$

$(\vec{r}-\vec{a}) \vec{n}_2=0 \Rightarrow y+z=-1$

$\qquad$.............$(2)$

Let the line be $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$ intersects the lines

$\Rightarrow \mathrm{S} . \mathrm{D}=0 \Rightarrow \mathrm{a}+3 \mathrm{~b}+5 \mathrm{c}=0 \text { and } 3 \mathrm{a}+\mathrm{b}-5 \mathrm{c}=0 \Rightarrow \mathrm{a}: \mathrm{b}: \mathrm{c}:: 5 \mathrm{r}:-5 \mathrm{r}: 2 \mathrm{r}$

on solving with given lines we get points of intersection $\mathrm{P} \equiv(5,-5,2)$ and $\mathrm{Q} \equiv\left(\frac{10}{3},-\frac{10}{3}, \frac{8}{3}\right)$ $\Rightarrow \mathrm{PQ}^2=\mathrm{d}^2=6$.

$(B)-(p), (r)$

$ \tan ^{-1}(\mathrm{x}+3)-\tan ^{-1}(\mathrm{x}-3)=\sin ^{-1}(3 / 5) $

$ \Rightarrow \tan ^{-1} \frac{(x+3)-(x-3)}{1+\left(x^2-9\right)}=\tan ^{-1} \frac{3}{4} \Rightarrow \frac{6}{x^2-8}=\frac{3}{4} $

$ \therefore \mathrm{x}^2-8=8 $

$ \text { or } \mathrm{x}= \pm 4 \text {. }$

$(C)-(q), (s)$

As $\bar{a}=\mu \vec{b}+4 \vec{c} \Rightarrow \mu(|\vec{b}|)=-4 \vec{b} \cdot \vec{c}$ and $|\vec{b}|^2=4 \vec{a} \cdot \vec{c}$ and $|\vec{b}|^2+\vec{b} \cdot \vec{c}-\vec{d} \cdot \vec{c}=0$ Again, as $2|\vec{b}+\vec{c}|=|\vec{b}-\vec{a}|$

Solving and eliminating $\vec{b} \cdot \vec{c}$ and eliminating $|\vec{a}|^2$ we get $\left(2 \mu^2-10 \mu\right)|\vec{b}|^2=0 \Rightarrow \mu=0$ and $5$ .

$(D)-(r)$

$ \mathrm{I}=\frac{2}{\pi} \int_{-\pi}^\pi \frac{\sin 9(x / 2)}{\sin (x / 2)} d x=\frac{2}{\pi} \times 2 \int_0^\pi \frac{\sin 9(x / 2)}{\sin (x / 2)} d x $

$ \mathrm{x} / 2=\theta \Rightarrow \mathrm{dx}=2 \mathrm{~d} \theta $

$ \mathrm{x}=0, \theta=0 $

$ \mathrm{x}=\pi \theta=\pi / 2 $

$ \mathrm{I}=\frac{8}{\pi} \int_0^{\pi / 2} \frac{\sin 9 \theta}{\sin \theta} d \theta $

$ =\frac{8}{\pi} \int_0^{\pi / 2} \frac{(\sin 9 \theta-\sin 7 \theta)}{\sin \theta}+\frac{(\sin 7 \theta-\sin 5 \theta)}{\sin \theta}+\frac{(\sin 5 \theta-\sin 3 \theta)}{\sin \theta}+\frac{(\sin 3 \theta-\sin \theta)}{\sin \theta}+\frac{\sin \theta}{\sin \theta} d \theta$

$ =\frac{16}{\pi} \int_0^{\pi / 2}(\cos 8 \theta+\cos 6 \theta+\cos 4 \theta+\cos 2 \theta+1) d \theta+\frac{8}{\pi} \int_0^{\pi / 2} d \theta $

$ =\frac{16}{\pi}\left[\frac{\sin 8 \theta}{8}+\frac{\sin 6 \theta}{6}+\frac{\sin 4 \theta}{4}+\frac{\sin 2 \theta}{2}\right]+\frac{8}{\pi}[\theta]_0^{\pi / 2}=0+\frac{8}{\pi} \times\left[\frac{\pi}{2}-0\right]=4$

$(2\lambda + 1,\,3\lambda - 1,\,4\lambda + 1);\,\,\lambda \in R$

Any point on $\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$ = \mu $ is,

$(\mu + 3,\,2\mu + k,\,\mu );\,\mu \in R$

The given lines intersect if and only if the system of equations (in $\lambda \,$ and $\mu $)

$2\lambda + 1 = \mu + 3.....(i)$

$3\lambda - 1 = 2\mu + k.....(ii)$

$4\lambda + 1 = \mu .....(iii)$

has a unique solution.

Solving $(i)$ and $(iii),$

we get $\lambda = \frac{{ - 3}}{2},\,\mu = - 5$

From $(ii)$, we get $\frac{{ - 9}}{2} - 1 = - 10 + k$

$\Rightarrow k = \frac{9}{2}$.

$b = \sqrt {1 + 4 + 4} = \sqrt 9 $

and $c = \sqrt {1 + 16 + 1} = \sqrt {18} $.

Obviously it is a right angled and isosceles triangle.

$\therefore $ Direction cosines are, $\left( {\frac{2}{{\sqrt {49} }},\,\frac{3}{{\sqrt {49} }},\, - \frac{6}{{\sqrt {49} }}} \right)$

or $\left( {\frac{2}{7},\,\frac{3}{7},\,\frac{{ - 6}}{7}} \right)$.

$z = \frac{{ - 5\,(3) + 3\,(6)}}{{ - 2}} = - \frac{3}{2}$.

and from $y$ - axis $ = \sqrt {1 + 9} = \sqrt {10} $.

$\, \Rightarrow \,\Sigma {\sin ^2}\alpha = 3 - 1 = 2$.

$ \Rightarrow \,\,\cos \,\gamma = \sqrt {1 - {{\left( {\frac{{14}}{{15}}} \right)}^2} - {{\left( {\frac{1}{3}} \right)}^2}} $

$ = \sqrt {\frac{8}{9} - \left( {\frac{{196}}{{225}}} \right)} = \pm \frac{2}{{15}}$.

From $y$-axis $ = \sqrt {1 + 9} = \sqrt {10} $

From $z$-axis $ = \sqrt {1 + 4} = \sqrt 5 $.

$\Rightarrow \,\,a = - 2,\,\,0 = \frac{{1 + b + 7}}{3}\,$

$\Rightarrow \,\,b = - \,8$

and $0 = \frac{{3 - 5 + c}}{3}\,\,$

$\Rightarrow \,\,c = 2$.

$\frac{{4 - ( - 2)}}{{ - 3 - 4}} \ne \frac{{ - 3 - 4}}{{ - 2 - ( - 3)}}$

Therefore, this set of points is non-collinear.

which is not possible.

or $\frac{2}{7},\,\frac{{ - 3}}{7},\,\,\frac{6}{7}$.

we get that neither $PQ\,\,||\,\,RS$ nor $PQ\,\, \bot \,\,RS.$ also $PQ \ne RS.$

therefore projection of $AB$ on $CD$ is $0$.

so $OP \bot \,OQ$.

$\Rightarrow \,\,\lambda = - \frac{1}{6}$

$\therefore \,\,x = \frac{{ - \,5 + 18}}{5} = \frac{{13}}{5},\,\,\,y = \frac{{ - 1 + 24}}{5} = \frac{{23}}{5}$.

${z_2} - {z_1} = 4$ and d.c's of $x,\,y,\,z$ - axes are $(1,0,0), (0,1,0), (0, 0, 1)$ respectively.

Now projection = $({x_2} - {x_1})l + ({y_2} - {y_1})m + ({z_2} - {z_1})n$

$\therefore $ Projections of line $AB$ on co-ordinate axes are $6, 6, 4$ respectively.

==> $ - \frac{5}{2} = - \frac{5}{2} = - \frac{5}{2}$ Obviously, points are collinear.

$\therefore \,\,5 = \frac{{9\lambda + 3}}{{\lambda + 1}}\,\,$

$\Rightarrow \,\,4\lambda = 2\,\, $

$\Rightarrow \,\,\lambda = \frac{1}{2}$

Hence required ratio is $2 : 1$.

or $\left( {\frac{2}{7},\frac{3}{7},\frac{6}{7}} \right)$.

$ \Rightarrow \,\,a = 4,\,\,b = 3,\,\,c = - 5$.

 Required co-ordinates of the points are,

$\,\,\left[ {\frac{{6 - 6}}{5},\frac{{10 + 3}}{5},\frac{{ - 14 + 24}}{5}} \right] = \left( {0,\,\frac{{13}}{5},\,\,2} \right)$.

= ${(x - a)^2} + {y^2} + {z^2} = {x^2} + {(y - b)^2} + {z^2} = {x^2} + {y^2} + {(z - c)^2}$

Therefore $x = \frac{a}{2},\,\,y = \frac{b}{2}$ and $z = \frac{c}{2}$.

= $[2 - ( - 1)]\frac{6}{7} + [5 - 0]\frac{2}{7} + [1 - 3]\frac{6}{7}$

= $\frac{{18 + 10 - 6}}{7}$= $\frac{{22}}{7}$.

$B = ( - 2,\,4,\,1)$;

$C = ( - 1,\,5,5)$

$D = (2,\,2,\,5)$

$AB = \sqrt {9 + 9 + 0} = 3\sqrt 2 ,\,\,$

$BC = \sqrt {1 + 1 + 16} = 3\sqrt 2 $

and $CD = 3\sqrt 2 $

and $AD = 3\,\sqrt 2 .$

Hence it is a square.

and ${l_2}x + {m_2}y + {n_2}z + d = 0$ .....$(ii)$

If $lx + my + nz + d = 0$ is perpendicular to $(i)$ and $(ii),$

then, $l{l_1} + m{m_1} +n{n_1} = 0,\,\,l{l_2} + m{m_2} + n{n_2} = 0\,$

$ \Rightarrow {\rm{ }}\,\frac{l}{{{m_1}{n_2} - {m_2}{n_1}}} = \frac{m}{{{n_1}{l_2} - {l_1}{n_2}}} = \frac{n}{{{l_1}{m_2} - {l_2}{m_1}}} = d$

Therefore, direction cosines are

$({m_1}{n_2} - {m_2}{n_1}),\,\,({n_1}{l_2} - {l_1}{n_2}),\,({l_1}{m_2} - {l_2}{m_1})$.

$ = 2\,{\cos ^2}\alpha - 1 + 2\,{\cos ^2}\beta - 1 + 2\,{\cos ^2}\gamma - 1$

$ = 2\,({\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma ) - 3 = 2 - 3 = - 1.$

$\therefore \,\,\,\frac{{3\lambda - 1}}{{\lambda + 1}} = 0\,\, $

$\Rightarrow \,\,\lambda = \frac{1}{3}$.

$\frac{3}{{\sqrt {{3^2} + {{12}^2} + {4^2}} }},\,\,\frac{{12}}{{\sqrt {{3^2} + {{12}^2} + {4^2}} }},\,\,\frac{4}{{\sqrt {{3^2} + {{12}^2} + {4^2}} }}$

$i.e.$, $\frac{3}{{13}},\frac{{12}}{{13}},\frac{4}{{13}}$.

$\frac{1}{{\sqrt {{1^2} + {2^2} + {3^2}} }},\frac{2}{{\sqrt {{1^2} + {2^2} + {3^2}} }},\frac{{ - \,3}}{{\sqrt {{1^2} + {2^2} + {3^2}} }}$

$i.e.$, $\frac{1}{{\sqrt {14} }},\frac{2}{{\sqrt {14} }},\frac{{ - \,3}}{{\sqrt {14} }}$.

But $x + 2y - 3z + 4 = 0$ can be written as $ - x - 2y + 3z - 4 = 0$.

Thus the direction cosines are $\frac{{ - 1}}{{\sqrt {14} }},\frac{{ - 2}}{{\sqrt {14} }},\frac{3}{{\sqrt {14} }}$.

So direction cosine are,$\left( {\frac{{ - 1}}{{\sqrt 6 }},\,\frac{2}{{\sqrt 6 }},\frac{{ - 1}}{{\sqrt 6 }}} \right)$.

$z = \frac{{{m_1}{z_2} + {m_2}{z_1}}}{{{m_1} + {m_2}}}$;

$x = \frac{{20}}{7},\,\,y = \frac{5}{7},\,\,z = \frac{{15}}{7}.$