Let $<a,b,c>$ be direction ratios of plane containing the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$ and $\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$

Since plane normal will be perpendicular to given lines then by condition of perpendicular lines, 

$2a+3b+5c=0 ....\left(1\right)$ and $3a+7b+8c=0.....\left(2\right)$ 

On solving $\left(1\right) \& \left(2\right)$ we get,

$\frac{a}{24-35}=\frac{b}{15-16}=\frac{c}{14-9}$

So, DR of plane are$<11,1,-5>$

Now let DR of plane $P$ be $<a_1,b_1,c_1>$

Then, $11a_1+b_1-5c_1=0 ....\left(3\right)$

And $9a_1-b_1-5c_1=0......\left(4\right)$ (as plane normal will be perpendicular to given line $\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$)

Now solving equation $\left(3\right) \& \left(4\right)$ we get, direction ratio $\left(1,-1,2\right)$

So, equation of plane will be $x-y+2z=21$

Now finding distance of point $\left(2,-5,11\right)$ from plane $x-y+2z=21$

$d=\left|\frac{2+5+22-21}{\sqrt{1^2+1^2+2^2}}\right|$

 $\Rightarrow d=\frac{8}{\sqrt{6}}$

$\Rightarrow d^2=\frac{32}{3}$