Let P be the plane containing the straight line (x-3)/9=(y+4)/-1=(z-7)/-5 and perpendicular to the plane containing the straight lines x/2=y/3=z/5…

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