Let $Q$ be the foot of the perpendicular from the point $P(7, -2, 13)$ on the plane containing the lines
$ \frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8} \quad \text{and} \quad \frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}. $
Then $(PQ)^2$, is equal to ________.
