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Let Q be the mirror image of the point P $\mathrm{\left ( 1,0,1 \right )}$ with respect to the plane $S: x+y+z=5 .$ If a line L passing through $\mathrm{\left ( 1,-1,-1 \right )}$ , parallel to the line PQ meets the plane S at R, then $QR^{2}$ is equal to:
Option: 1
$2$

Option: 2
$5$

Option: 3
$7$

Option: 4
$11$

Answers (1)

best_answer

By symmetry, QR = PR
The eqn of the plane $\mathrm{S:\, x+y+z=5---(1)}$
Eqn of the line L passing through $\mathrm{(1,-1,-1)}$ and parallel to the line PQ
$\mathrm{\Rightarrow L: \frac{x-1}{1}=\frac{y+1}{1}=\frac{z+1}{1}=r^{\prime}}$
General point $\mathrm{R\left [ r^{\prime}+1,r^{\prime}-1,r^{\prime}-1 \right ]}$

$\mathrm{\because \, \text{Point R lies on the plane (1)}}$
$\mathrm{\therefore \, r^{\prime}+1+r^{\prime}-1+r^{\prime}-1=5}$
$\mathrm{ 3r^{\prime}= 6\: \Rightarrow \;r^{\prime}= 2 }$

$\mathrm{ R= \left ( 3,1,1 \right ) \;}$

$\mathrm{ Q R^{2}=PR^2 }$
$\mathrm{ =\left ( 4+1+0 \right )= 5 }$

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SANGALDEEP SINGH

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