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

$$\begin{gathered} \begin{array}{l}\mathrm{a}(\mathrm{x}-3)+\mathrm{b}(\mathrm{y}+4)+\mathrm{c}(\mathrm{z}-7)=0\\ \mathrm{P}:9\mathrm{a}-\mathrm{b}-5\mathrm{c}=0\\ drs of line perpendicular to the given lines=\left|\begin{array}{ccc}i& j& k\\ 2& 3& 5\\ 3& 7& 8\end{array}\right|\\ \end{array} \\ =(-11,-1,5) \\ \Rightarrow -11\mathrm{a}-\mathrm{b}+5\mathrm{c}=0 \end{gathered}$$

After solving DR’s (a,b,c)  $\propto (1,-1,2)$

Equation of plane

$\begin{array}{l}x-y+2z=21\\ Perpendicular dis\tan ce from (2,-5,11)\\ =\frac{\left|2+5+22-21\right|}{\sqrt{1+1+4}}\\ d=\frac{8}{\sqrt{6}}\\ d^2=\frac{32}{3}\end{array}$