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

Equation of the plane

$A(x+1)+B(y-1)+C(z-3)=0$

where $6 \mathrm{A}+7 \mathrm{B}+8\mathrm{C}=0$ and $3 \mathrm{A}+5 \mathrm{B}+7\mathrm{C}=0$

$$By cross multiplication method, we get

$\frac{A}{1}=\frac{B}{-2}=\frac{C}{1}$

$\Rightarrow 1(x+1)-2(y-1)+1(z-3)=0\Rightarrow x-2y+z=0$

Let $Q(\alpha ,\beta ,\gamma )$

$\frac{\alpha -7}{1}=\frac{\beta +2}{-2}=\frac{\gamma -13}{1}=\frac{-(7+4+13)}{1+4+1}=-4$

$\mathrm{Q}(\alpha ,\beta ,\gamma )=Q(3,6,9)$

$PQ=\sqrt{{\left(7-3\right)}^2+{\left(-2-6\right)}^2+{\left(13-9\right)}^2}$

$(PQ{)}^2=16+64+16$

$(PQ{)}^2=96$