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Let $\mathrm{P}$ be the plane containing the traight line $\mathrm{\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}}$ and perpendicular to the plane containing the straight lines $\mathrm{\frac{x}{2}=\frac{y}{3}=\frac{z}{5} \text { and } \frac{x}{3}=\frac{y}{7}=\frac{z}{8}}$ . If $\mathrm{d}$ is thed distance of $\mathrm{P}$ from the point $\mathrm{(2,-5,11)}$ , then $\mathrm{d^{2}}$ is equal to :
Option: 1
$\frac{147}{2}$

Option: 2
$96$

Option: 3
$\frac{32}{3}$

Option: 4
$54$

Answers (1)

best_answer

$\mathrm{a(x-3)+b(y+4)+c(z-7)=0} \\$

$\mathrm{p: 9 a-b-5 c=0} \\$

$\mathrm{-11 a-b+5 c=0}$

After solving DR's $\mathrm{\alpha(1,-1,2)}$

Equation of plane

$\mathrm{x-y+2 z=21} \\$

$\mathrm{d=\frac{8}{\sqrt{6}}} \\$

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

Hence correct option is 3

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