Let $\mathrm{P}$ be a plane passing through the points $(2,1,0),(4,1,1)$ and $(5,0,1)$ and $R$ be the point…
Let $\mathrm{P}$ be a plane passing through the points $(2,1,0),(4,1,1)$ and $(5,0,1)$ and $R$ be the point $(2,1,6)$. Then image of $\mathrm{R}$ in the plane $\mathrm{P}$ is
$(6,5,2)$
$(4,3,2)$
$(6,5,-2)$
$(3,4,-2)$
Solution
Equation of the plane passing through $(2,1,0)$, $(4,1,1)$ and $(5,0,1)$ is
$\begin{aligned}
& \left|\begin{array}{lll}
x-2 & y-1 & z-0 \\
4-2 & 1-1 & 1-0 \\
5-2 & 0-1 & 1-0
\end{array}\right|=0 \\
& \Rightarrow x+y-2 z=3
\end{aligned}$
$\mathrm{R}^{\prime}(x, y, \mathrm{z})$ is image of $\mathrm{R}(2,1,6)$ w:r.t. to plane
$\begin{aligned}
& x+y-2 z=3 \\
& \frac{x-2}{1}=\frac{y-1}{1}=\frac{z-6}{-2}=\frac{-2[2+1-2(6)-3]}{1+1+4} \\
& \Rightarrow \frac{x-2}{1}=\frac{y-1}{1}=\frac{z-6}{-2}=4 \\
& \Rightarrow x=6, y=5, \mathrm{z}=-2 \\
\therefore \quad & \mathrm{R}^{\prime}(x, y, \mathrm{z}) \equiv(6,5,-2)
\end{aligned}$