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(C) The equation of the plane passing through $(2,1,0)$,$(4,1,1)$,and $(5,0,1)$ is given by the determinant equation: $\begin{vmatrix} x-2 & y-1 & z-0

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$(6,5,-2)$

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(C) The equation of the plane passing through $(2,1,0)$,$(4,1,1)$,and $(5,0,1)$ is given by the determinant equation: $\begin{vmatrix} x-2 & y-1 & z-0 \ 4-2 & 1-1 & 1-0 \ 5-2 & 0-1 & 1-0 \end{vmatrix} = 0$ $\Rightarrow \begin{vmatrix} x-2 & y-1 & z \ 2 & 0 & 1 \ 3 & -1 & 1 \end{vmatrix} = 0$ Expanding the determinant: $(x-2)(0+1) - (y-1)(2-3) + z(-2-0) = 0$ $\Rightarrow (x-2) + (y-1) - 2z = 0 \Rightarrow x+y-2z = 3$. Let $R'(x, y, z)$ be the image of $R(2,1,6)$ with respect to the plane $x+y-2z-3=0$. The formula for the image $(x, y, z)$ of a point $(x_1, y_1, z_1)$ in the plane $ax+by+cz+d=0$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -2 \frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$. Substituting the values: $\frac{x-2}{1} = \frac{y-1}{1} = \frac{z-6}{-2} = -2 \frac{2+1-2(6)-3}{1^2+1^2+(-2)^2} = -2 \frac{3-12-3}{6} = -2 \frac{-12}{6} = 4$. Thus,$x-2 = 4 \Rightarrow x=6$,$y-1 = 4 \Rightarrow y=5$,and $z-6 = -8 \Rightarrow z=-2$. Therefore,the image $R'$ is $(6, 5, -2)$.

Let $P$ 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 the image of $R$ in the plane $P$ is

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