JEE Main 2017 — Three Dimensional Geometry Question with Solution

Any line parallel to $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ and passing through $P(1,-2,3)$ is

$\frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{5}=\lambda$ (say)

Any point on above line can be written as $(\lambda +1,4\lambda -2,5\lambda +3).$

$\therefore$ Coordinates of $R$ are $(\lambda +1,4\lambda -2,5\lambda +3).$

Since, point $R$ lies on the above plane. $\therefore 2(\lambda +1)+3(4\lambda -2)-4(5\lambda +3)+22=0$

$\Rightarrow \lambda =1$

So, point $R$  is $(2,2,8)$

Now, $PR=\sqrt{(2-1{)}^2+(2+2{)}^2+(8-3{)}^2}=\sqrt{42}$

$\therefore PQ=2PR=2\sqrt{42}$

Since, point $R$ lies on the above plane.

So, point $R$ is $(2,2,8)$