JEE Main 2018 — Three Dimensional Geometry Question with Solution

Since the plane bisects the line joining the points $(1,2,3)$ and $(-3,4,5)$ then the plane passes through the midpoint of the line which is : $$\begin{array}{rl}\left(\frac{1-3}{2},\frac{2+4}{2},\frac{5+3}{2}\right)\equiv (\frac{-2}{2},& \frac{6}{2},\frac{8}{2})\\ & \equiv (-1,3,4).\end{array}$$ As plane cuts the line segment at right angle, so the direction cosines of the normal of the plane are $(-3-1,4-2,5-3)=(-4,2,2)$ So the equation of the plane is : $$-4x+2y+2z=\lambda$$ As plane passes through $(-1,3,4)$ so $$-4(-1)+2(3)+2(4)=\lambda \Rightarrow \lambda =18$$ Therefore, equation of plane is : $$-4x+2y+2z=18$$ Now, only $(-3,2,1)$ satisfies the given plane as $-4(-3)+2(2)+2(1)=18$