Let $A$ be the point of intersection of the lines $L_1: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}$ and $L_2: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$. Let $B$ and $C$ be the points on the lines $L_1$ and $L_2$ respectively such that $A B=A C=\sqrt{15}$. Then the square of the area of the triangle ABC is :