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A plane P contains the line of intersection of the plane $\mathrm{\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k}) = 6}$ and $\mathrm{\overrightarrow{r}.(\hat{2i}+\hat{3j}+\hat{4k}) =-5}$ . If P passes
through the point $(0, 2, -2),$ then the square of distance of the point $(12, 12, 18)$ from the plane P is :
Option: 1
620

Option: 2
1240

Option: 3
310

Option: 4
155

Answers (1)

best_answer

eqⁿof plane $\mathrm{P_{1}+\lambda P_{2}=0}$

$\mathrm{(x+y+z-6)+\lambda(2 x+3 y+4 z+5)=0 }$

pass th. $(0,2,-2)$

$\mathrm{(-6)+\lambda(6-8+5)=0 }$

$\mathrm{(-6)+\lambda[3]=0}$ $\Rightarrow \lambda=2 \\$

eqⁿof plane

$\mathrm{5x+7y+9z+4=0}$

distance from (12,12,18)

$\mathrm{ \mathrm{d}=\left|\frac{60+84+162+4}{\sqrt{25+49+81}}\right| }$

$\mathrm{\mathrm{d}=\frac{310}{\sqrt{155}} }$

$\mathrm{d^2=\frac{310 \times 310}{155} }$

$\mathrm{ \mathrm{~d}^2=620}$

Correct answer is option 1

Posted by

Ritika Jonwal

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