Let 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 d is the distance P from the point (2, –5, 11), then d2 is equal to :