Given equation of circle is
$x^{2}+y^{2}=169$
Its centre $=(0,0)$ and radius $=13$
Now, slope of $O R=\frac{-5}{12}=m_{1}\left(\because\right.$ slope $\left.=\frac{y_{2}-y}{x_{2}-x}\right)$
and slope of $O Q=\frac{12}{5}=m_{2}$
$\because m_{1} \cdot m_{2}=-1 $
$\Rightarrow \angle R O Q=\frac{\pi}{2}$
We know that, angle made by the chord of circle at circumference is equal to the half of the angle made by the same chord at the centre of circle.
$\therefore \angle Q P R=\frac{1}{2} \cdot \angle R O Q=\frac{1}{2} \times \frac{\pi}{2}=\frac{\pi}{4}$
