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 :
1
\(\frac{147}{2}\)
2
96
3
\(\frac{32}{3}\)
4
54