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Let the image of the point P(2,-1,3) in the plane x+2 y-z=0 be Q. Then the distance of the plane3 x+2 y+z+29=0 from the point Q is

Option: 1

\frac{24 \sqrt{2}}{7}


Option: 2

2 \sqrt{14}


Option: 3

3 \sqrt{14}


Option: 4

\frac{22 \sqrt{2}}{7}


Answers (1)

best_answer

let \mathrm{Q}(\alpha, \beta, \gamma) is image of \mathrm{P}(2,-1,3) in the plane \mathrm{x}+2 \mathrm{y}-\mathrm{z}=0
$$ \begin{aligned} & \frac{\alpha-2}{1}=\frac{\beta+1}{2}=\frac{\gamma-3}{-1}=\frac{-2(2-2-3)}{1^2+2^2+(-1)^2}=1 \\ & \alpha=3, \beta=1, \gamma=2 \end{aligned}
Distance of  \mathrm{Q}(3,1,2) from
$$ \begin{aligned} & 3 x+2 y+z+29=0 \\ & \mathrm{D}=\frac{|3(3)+2(1)+2+29|}{\sqrt{3^2+2^2+1^2}} \\ & =\frac{42}{\sqrt{14}}=3 \sqrt{14} \end{aligned}

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Shailly goel

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