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

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
\(\begin{aligned}
& x+y-2 z=3 \\
& \frac{x-2}{1}=\frac{y-1}{1}=\frac{z-6}{-2}=\frac{-2[2+1-2(6)-3]}{1+1+4} \\
& \Rightarrow \frac{x-2}{1}=\frac{y-1}{1}=\frac{z-6}{-2}=4 \\
& \Rightarrow x=6, y=5, \mathrm{z}=-2 \\
\therefore \quad & \mathrm{R}^{\prime}(x, y, \mathrm{z}) \equiv(6,5,-2)
\end{aligned}\)