Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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