Let $\mathrm{P}$ be the image of $\mathrm{O}$ in the given plane.


Equation of the plane, $4 x-3 y+z+13=0$ 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
$$
\begin{aligned}
& \frac{x-0}{4}=\frac{y-0}{-3}=\frac{z-0}{1}=r \text { (let) } \\
\therefore \quad & x=4 r, y=-3 r, z=r
\end{aligned}
$$
Let the coordinate of $\mathrm{P}$ be $(4 r,-3 r, r)$
Since $Q$ be the mid point of OP
$$
\therefore \quad \mathrm{Q}=\left(2 r,-\frac{3}{2} r, \frac{r}{2}\right)
$$
Since Q lies in the given plane
$$
\begin{aligned}
& 4 x-3 y+z+13=0 \\
\therefore & 8 r+\frac{9}{2} r+\frac{r}{2}+13=0 \\
\Rightarrow & r=\frac{-13}{8+\frac{9}{2}+\frac{1}{2}}=\frac{-26}{26}=-1
\end{aligned}
$$
$\begin{aligned} \therefore \quad & \mathrm{Q}=\left(-2, \frac{3}{2},-\frac{1}{2}\right) \\ & \mathrm{QR}=\sqrt{(-1+2)^2+\left(1-\frac{3}{2}\right)^2+\left(-6+\frac{1}{2}\right)^2} \\ & =\sqrt{1+\frac{1}{4}+\frac{121}{4}}=3 \sqrt{\frac{7}{2}}\end{aligned}$