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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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$(6,5,-2)$

Vedclass · Answer

(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$$\Rightarrow \begin{vmatrix} x-2 & y-1 & z \ 2 & 0 & 1 \ 3 & -1 & 1 \end{vmatrix} = 0$Expanding along the first row:$(x-2)(0 - (-1)) - (y-1)(2 - 3) + z(-2 - 0) = 0$$(x-2)(1) - (y-1)(-1) - 2z = 0$$x - 2 + y - 1 - 2z = 0$$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}$$\frac{x-2}{1} = \frac{y-1}{1} = \frac{z-6}{-2} = -2 \frac{3 - 12 - 3}{1 + 1 + 4} = -2 \frac{-12}{6} = 4$Equating each part to $4$:$x - 2 = 4 \Rightarrow x = 6$$y - 1 = 4 \Rightarrow y = 5$$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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