Question:
Consider the line $L$ given by the equation $\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1} .$ Let $Q$ be the mirror image of the point $(2,3,-1)$ with respect to $L$. Let a plane $P$ be such that it passes through $Q$, and the line $L$ is perpendicular to P. Then which of the following points is on the plane $\mathrm{P}$ ?
Correct Option: , 4
Solution:
Plane $p$ is $\perp^{\mathrm{r}}$ to line
$\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1}$
\& passes through pt. $(2,3)$ equation of plane $p$
$2(x-2)+1(y-3)+1(z+1)=0$
$2 x+y+z-6=0$
pt $(1,2,2)$ satisfies above equation
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