JEE Main 2014 — Three Dimensional Geometry Question with Solution

We first find the intersection of line $\left(L_1\right)$ with the plane. Any general point on line can be written as $\left(3k+1,k+3,-5k+4\right)$.

Substituting in equation of plane $P$ we get

$2(3k+1)-(k+3)+(-5k+4)+3=0$
Which gives $6=0$ which is not possible.

So, we conclude that the line is parallel to the plane.$\left(\because 6-1-5=0\right)$

$\therefore$ The reflected line $\left(L_2\right)$ will also have the same direction ratios. i.e.,$(3,1,- 5).$
Consider the normal perpendicular to the plane and line.
Its direction ratios are $(2,-1,1)$

$\therefore$ Equation of normal is

$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}$

$\therefore$ Any general point can be written as $(2r+1,-r+3,r+4)$ substituting in the equation of a plane,

 $2(2r+1)-(-r+3)+\left(r +4\right)+3=0$

$4r+2+ r-3+r+4+3 = 0$

$\Rightarrow 6r+6=0$

$r=-1$

​

$\therefore$ The point $B$ is $(-1,4,3)$

$B$ is the midpoint of segment $AC$.

$C\equiv \left(-3,5,2\right)$

$\therefore$Equation of line $L_2$ is $\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$