3d Geometry

Let $\mathrm{P}$ be the plane containing the straight line $\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$ and perpendicular to the plane containing the straight lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$ and $\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$. If $\mathrm{d}$ is the distance of $\mathrm{P}$ from the point $(2,-5,11)$, then $\mathrm{d}^{2}$ is equal to :