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 :