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 :

Let $\langle a, b, c\rangle$ be direction ratios of plane containing <br/><br/> $$ \begin{aligned} &\text { lines } \frac{x}{2}=\frac{y}{3}=\frac{z}{5} \text { and } \frac{x}{3}=\frac{y}{7}=\frac{z}{8} . \\\\ &\therefore \quad 2 a+3 b+5 c=0 \quad \ldots \text { (i) } \\\\ &\text { and } 3 a+7 b+8 c=0 \quad \ldots \text { (ii) } \\\\ &\text { from eq. (i) and (ii) }: \frac{a}{24-35}=\frac{b}{15-16}=\frac{c}{14-9} \\\\ &\therefore \text { D. R's} \text {. of plane are }&lt;11,1,-5&gt; \end{aligned} $$ <br/><br/> Let $D . R$'s of plane $P$ be $< a_{1}, b_{1}, c_{1} >$ then. <br/><br/> $11 a_{1}+b_{1}-5 c_{1}=0$ <br/><br/> and $9 a_{1}-b_{1}-5 c_{1}=0$ <br/><br/> From eq. (iii) and (iv) : <br/><br/> $$ \begin{aligned} &\frac{a_{1}}{-5-5}=\frac{b_{1}}{-45+55}=\frac{c_{1}}{-11-9} \\\\ &\therefore \text { D.R's } \text { of plane } P \text { are }< 1,-1,2 > \end{aligned} $$ <br/><br/> Equation plane $P$ is : $1(x-3)-1(y+4)+2(z-7)=0$ <br/><br/> $\Rightarrow x-y+2 z-21=0$ <br/><br/> Distance from point $(2,-5,11)$ is $d=\frac{|2+5+22-2|}{\sqrt{6}}$<br/><br/> $\therefore d^{2}=\frac{32}{3}$