JEE Main 2013 — Three Dimensional Geometry Question with Solution

Let $\mathrm{P}$ be the image of $\mathrm{O}$ in the given plane.
Equation of the plane, $4x-3y+z+13=0$ OP is normal to the plane, therefore direction ratio of OP are proportional to $4,-3,1$ Since OP passes through $(0,0,0)$ and has direction ratio proportional to $4,-3,1$. Therefore equation of OP is $$\begin{array}{rl}& \frac{x-0}{4}=\frac{y-0}{-3}=\frac{z-0}{1}=r\text{ (let) }\\ \therefore & x=4r,y=-3r,z=r\end{array}$$ Let the coordinate of $\mathrm{P}$ be $(4r,-3r,r)$ Since $Q$ be the mid point of OP $$\therefore \mathrm{Q}=\left(2r,-\frac{3}{2}r,\frac{r}{2}\right)$$ Since Q lies in the given plane $$\begin{array}{rl}& 4x-3y+z+13=0\\ \therefore & 8r+\frac{9}{2}r+\frac{r}{2}+13=0\\ \Rightarrow & r=\frac{-13}{8+\frac{9}{2}+\frac{1}{2}}=\frac{-26}{26}=-1\end{array}$$ $\begin{array}{rl}\therefore & \mathrm{Q}=\left(-2,\frac{3}{2},-\frac{1}{2}\right)\\ & \mathrm{QR}=\sqrt{(-1+2{)}^2+{\left(1-\frac{3}{2}\right)}^2+{\left(-6+\frac{1}{2}\right)}^2}\\ & =\sqrt{1+\frac{1}{4}+\frac{121}{4}}=3\sqrt{\frac{7}{2}}\end{array}$