Mid-point of AB=$\begin{array}{l} = \left( {\dfrac{{3 + 1}}{2},\dfrac{{1 + 3}}{2},\dfrac{{4 + 6}}{2}} \right)\ = (2,2,5)\end{array}$
lies on the plane as it satisfies the equation of the plane
and DR s of AB $ = (2, - 2,2)$
DR s of normal to the plane $= (1, - 1,1)$
AB is the perpendicular bisector.
Hence, A is the image of 2 
 

 Given plane equation is $2{x}+3{y}-2{z}-k=0$

Equation of line through $A(3,-2,7)$ and $B(13,3,-8)$ is:

We know that the ratio in which the plane $ax+by+cz+d=0$ divides the line segment joining (${x _1},{y _1},{z _1}$) and (${x _2},{y _2},{z _2}$) is

Direction ratios of the line is $2i+3j+2k$
and normal of plane is along $4i-2j-k$
Now, $(2i+3j+2k).(4i-2j-k)=8-6-2=0$
Therefore, line is parallel to plane

Ans: A

Let the plane $r.(i-2j+3k)=17$ divide the line joining the points. 

$-2i+4j+7k$ and $2i-5j+8k$ in the ratio $t:1$ at the point $P.$

$\therefore P$ is $\displaystyle \dfrac { 3t-1 }{ t+1 } i+\dfrac { -5t+4 }{ t+1 } j+\dfrac { 8t+7 }{ t+1 } k.$

This lies on the given plane, 

$\displaystyle \therefore \dfrac { 3t-2 }{ t+1 } .1+\dfrac { -5t+4 }{ t+1 } \left( 2 \right) +\dfrac { 8t+7 }{ t+1 } \left( 3 \right) =17$

$\Rightarrow 3t-2+10t-8+24t+21=17t+17$

$\displaystyle \therefore 20t=17-21+10=6\Rightarrow =\dfrac { 6 }{ 20 } =\dfrac { 3 }{ 10 } $

$\therefore$ required ratio is $3:10$.

Given line is $\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } $

Equation of plane in cartesian form is, $x-2y+3z-17=0$.....$(1)$
Assume this plane $(1)$ divide the line segment joining the points $(-2,4,7)$ and $(3,-5,8)$ in $m:n$ ratio
Therefore,  $\dfrac{m}{n} = \dfrac{-2-2(4)+3(7)-17}{3-2(-5)+3(8)-17}= \dfrac{-3}{10} < 0$
Hence, plane (1) divides the given line segment externally $3:10$ 

Normal of the plane $\vec { r } \cdot \vec { n } =q$ is $\vec { n } $ and direction ratio of the line $\vec { r } =\vec { a } +\lambda \vec { b } $ is $\vec { b } $
Since, line lies in the plane, normal and direction ratios should be perpendicular.
Therefore, $\vec { b } \cdot \vec { n } =0$
Also position vector $\vec { a } $ should lie on plane $\vec { r } \cdot \vec { n } =q$
Therefore, $\vec { a } \cdot \vec { n } =q$

Ans: C

Equation of plane in cartesian form is, $x-2y+3z-13=0$ .....(1)$
Assume this plane (1) divide the line segment joining the points
$(2,-4,-7)$ and $(-3,5,-8)$ in $m:n$ ratio
Therefore,  $\dfrac{m}{n} = \dfrac{2-2(-4)+3(-7)-13}{-3-2(5)+3(-8)-13}= \dfrac{-12}{25} < 0$
Henc,e plane (1) divides the given line segment externally  $12:25$.

consider $P: x+2y-z=0$
direction ratio of normal of $P$ is $(1,2,-1)$

a) for $L _{1}: \dfrac {x-1}{1}=\dfrac {y}{-1}=\dfrac {z-5}{-1}$
Since, $(1,0,5)$ does not lie in $P$ 
Therefore, $L _{1}$ does not lie in $P$

b) for $L _{2}: x-y+z=2x+y-z=0$
direction ratio of $L _{2}$ is $(i-j+k) \times (2i+j-k)=3(j+k)$
since, $(i+2j-k).(3j+3k)=6-3=3 \neq =0$
Therefore, $L _{2}$ does not lie on $P$

c) for $L _{3}:\vec r=2\hat i-\hat j+4\hat k+\lambda (3\hat i+\hat j+5\hat k)$
Since, $(2,-1,4)$ does not lie in $P$ 
Therefore, $L _{3}$  does not lie in $P$

Ans: D

Let P point lie on a plane $2x+2y-2z-1=0 $and divide $(2, 1, 5), (3, 4, 3)$ in a ratio of  $\lambda :1$
by section formula
P=($\frac { 3\lambda +2 }{ \lambda +1 } ,\frac { 4\lambda +1 }{ \lambda +1 } ,\frac { 3\lambda +5 }{ \lambda +1 } $)
put co-ordinate of P in plane equation
2$\left(\cfrac { 3\lambda +2 }{ \lambda +1 } +\cfrac { 4\lambda +1 }{ \lambda +1 } -\cfrac { 3\lambda +5 }{ \lambda +1 } \right)$=1
by solving above equation,
we get,
 $\lambda=\dfrac{5}{7}$

Given line is, $\dfrac{x-2}{2} = \dfrac{y+1}{4} = \dfrac{z-12}{2} = k$ (say)

Clearly direction ratios of perpendicular drawn from origin to the given plane are $3,4,-6$
Hence equation of perpendicular line to the given plane and  passing through origin is given by,
$\cfrac{x}{3}=\cfrac{y}{4}=\cfrac{z}{-6}=k$ (say)
Now let foot of perpendicular be $P(3k, 4k, -6k)$
Also this point lie in the given plane $\Rightarrow 3(3k)+4(4k)-6(-6k)+1=0\Rightarrow k = -\cfrac{1}{61}$
Hence $P \equiv \left(-\cfrac{3}{61}, -\cfrac{4}{61}, \cfrac{6}{61}\right)$

We know that the equation of the line passing through the point $(x _1,y _1,z _1) , (x _2,y _2,z _2) $

The line $\dfrac{x- \alpha^1}{l}=\dfrac{y-\beta^1}{m}=\dfrac{z-\gamma^1}{n}$ in the plane $Ax+By+Cz+D=0$

${ P } _{ 1 }\equiv x-2y+z=1$

Equation of the line is $\displaystyle \frac { x-2 }{ 2-3 } =\frac { y+3 }{ -3+4 } =\frac { z-1 }{ 1+5 } $ or $\displaystyle \frac { x-2 }{ -1 } =\frac { y+3 }{ 1 } =\frac { z-1 }{ 6 } $
Any point on the line $\left( -r+2,r-3,6r+1 \right) $ which cuts the $yz$-plane at the point where $-r+2=0$ or $r=2$
and the point of intersection is $\left( 0,-1,13 \right) $ 
Similarly it cuts the $zx$-plane at $\left( -1,0,19 \right) $ and $xy$ plane at $\displaystyle \left( \frac { 13 }{ 6 } ,\frac { -19 }{ 6 } ,0 \right) $.

Given line L: $\displaystyle \frac{x-1}{2} = \frac{y - 1}{1} = \frac{z - 0}{4} $

The D.C. of the line are $\begin{pmatrix}\displaystyle\frac{1}{\sqrt{2}},\,\displaystyle\frac{1}{\sqrt{2}},\,0\end{pmatrix}$
any point on the line at a distance $\lambda$ from $A(0,\,1,\,2)$
is $\begin{pmatrix}0+\displaystyle\frac{\lambda}{\sqrt{2}},\,1+\displaystyle\frac{\lambda}{\sqrt{2}},\,2+\lambda.0\end{pmatrix}$
which lies on $x+y+z=0$
$\therefore\;\displaystyle\frac{\lambda}{\sqrt{2}}+\begin{pmatrix}1+\displaystyle\frac{\lambda}{\sqrt{2}}\end{pmatrix}+2=0$
$\Rightarrow\;\displaystyle\frac{2\lambda}{\sqrt{2}}=-3\;\;\;\;\Rightarrow\;\;\;\;\lambda=\displaystyle\frac{-3}{\sqrt{2}}$
$\therefore\;B=\begin{pmatrix}-\displaystyle\frac{3}{2},\,-\displaystyle\frac{1}{2},\,2\end{pmatrix}$
$A=(0,\,1,\,2)\;&amp;\;B\equiv\;\begin{pmatrix}-\displaystyle\frac{3}{2},\,-\displaystyle\frac{1}{2},\,2\end{pmatrix}$
$AB=\sqrt{\displaystyle\frac{9}{4}+\displaystyle\frac{9}{4}}=\displaystyle\frac{3\sqrt{2}}{2}=\displaystyle\frac{3}{\sqrt{2}}$

Equation of plane in cartesian form is, $x-2y+3z-17=0.....(1)$
Assume this plane (1) divide the line segment joining the points $(-2,4,7)$ and $(3,-5,8)$ in $m:n$ ratio.
Therefore,  $\dfrac{m}{n} = \dfrac{-2-2(4)+3(7)-17}{3-2(-5)+3(8)-17}= \dfrac{-3}{10} < 0$
Hence plane (1) divides the given line segment externally  $3:10$.  

A straight line $\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } $ meets the plane $\overrightarrow { r } .\overrightarrow { n } =0$ at $P$ for which $\lambda$ is given by,

$\displaystyle \left( \overrightarrow { a } +\lambda \overrightarrow { b }  \right) .\overrightarrow { n } =0\Rightarrow \lambda =-\dfrac { \overrightarrow { a } .\overrightarrow { n }  }{ \overrightarrow { b } .\overrightarrow { n }  } $

Thus, the position vector of $P$ is

$\displaystyle \overrightarrow { r } =\overrightarrow { a } -\dfrac { \overrightarrow { a } .\overrightarrow { n }  }{ \overrightarrow { b } .\overrightarrow { n }  } \overrightarrow { b } $  $[$ putting the value of $\lambda$ in $\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } ]$

Given equation of straight line $\displaystyle \dfrac { x-4 }{ 1 } =\dfrac { y-2 }{ 1 } =\dfrac { z-k }{ 2 } $

Given line is $\displaystyle\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ any point on the line is $(t, 2t, 3t)$. It lies in the given plane, if $t-2(2t)+3t-6=0$
i.e., $0\cdot t=6$, which is not true for any real $t$. So, the line and plane do not meet. i.e., the line is parallel to the plane.

We have,

$\dfrac{x}{2}+\dfrac{y}{3}+\dfrac{z}{4}=1$

Compare this equation of plane,

$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$

Then,

$a=2,\,b=3,\,c=4$

Then,

The coordinate of X-axis is $=A\left( a,0,0 \right)$$=A\left( 2,0,0 \right)$

The coordinate of Y-axis is $=B\left( 0,b,0 \right)$$=A\left( 0,3,0 \right)$

The coordinate of X-axis is $=C\left( 0,0,c \right)$$=C\left( 0,0,2 \right)$

We know that, the area of  \[\Delta ABC\] is

$ Area\,of\,\Delta ABC=\dfrac{1}{2}\sqrt{{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}} $

$ =\dfrac{1}{2}\sqrt{{{2}^{2}}\times {{3}^{2}}+{{3}^{2}}\times {{4}^{2}}+{{4}^{2}}\times {{2}^{2}}} $

$ =\dfrac{1}{2}\sqrt{36+144+64} $

$ =\dfrac{1}{2}\sqrt{180+64} $

$ =\dfrac{1}{2}\sqrt{244} $

$ =\dfrac{1}{2}\sqrt{2\times 2\times 61} $

$ =\sqrt{61} $

Hence, this is the answer.

We have point $(0,3,4)$ and we have direction's ratio also of line so we can write equation of line
$\dfrac{x-0}{2}=\dfrac{y-3}{-2}=\dfrac{z-4}{1}=k$
So consider a point which is lied on line $(2k,-2k+3,k+4)$
This point lies on plane also so it will satisfy the equation of plane:
$\Rightarrow 2(2k)-2(-2k+3)+k+4-10=0$
$\Rightarrow k=\dfrac{4}{3}$
Substitute the value of $k$ in point $\left (\dfrac{8}{3},\dfrac{1}{3},\dfrac{16}{3}\right)$

The line is $\displaystyle \frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}2{}$

Given  lines
 $x -2y + 4z + 4 = 0$    ....(1)
 $x + y + z - 8 = 0$       .....(2)
Subtracting (2) from (1), we get
$\Rightarrow y-z=4$        .....(3)
Given equation of plane $x-y+2z+1=0$
Since, the line intersects the plane, so using (3), we get 
$\Rightarrow x+z=3$      ......(4)
Hence, $y=5$, $z=1 $ and $x=2$
Hence, option C is correct.

Equation of a line through the point $\displaystyle P\left ( 2,-1,2 \right )$, equally inclined to the axes is given by,
$\displaystyle \frac{x-2}{1}=\frac{y+1}{1}=\frac{z-2}{1}=k$ (say)
Any point on the line is $Q ( k+2,k-1, k+2  )$ which lies on the plane $\displaystyle 2x+y+z=9$
 $\Rightarrow \displaystyle 2\left ( k+2 \right )+k-1+k+2=9 \Rightarrow k=1$,
For this values of $k$, the coordinates of $Q$ are $\displaystyle \left ( 3,0,3 \right )$.
So, $\displaystyle PQ=\sqrt{\left ( 3-2 \right)^2+\left ( 0+1 \right )^2+\left ( 3-2 \right )^2}=\sqrt{3}$

Equation of line passing through $(5, 1, a)$ and $(3, b, 1)$ is
$\dfrac{x-3}{5-3}= \dfrac{y-b}{1-b}= \dfrac{z-1}{a-1}$   ...(i)