Let Q be the mirror image of the point P(1, 0 | vedclass

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(B) The mirror image $Q(a, b, c)$ of point $P(1, 0, 1)$ with respect to the plane $x + y + z - 5 = 0$ is given by the formula $\frac{a-1}{1} = \frac{b

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$5$

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(B) The mirror image $Q(a, b, c)$ of point $P(1, 0, 1)$ with respect to the plane $x + y + z - 5 = 0$ is given by the formula $\frac{a-1}{1} = \frac{b-0}{1} = \frac{c-1}{1} = -2 \frac{1(1) + 1(0) + 1(1) - 5}{1^2 + 1^2 + 1^2} = -2 \frac{-3}{3} = 2$.Thus,$a-1 = 2 \Rightarrow a = 3$,$b = 2$,$c-1 = 2 \Rightarrow c = 3$. So,$Q = (3, 2, 3)$.The vector $\vec{PQ} = (3-1, 2-0, 3-1) = (2, 2, 2)$,which is parallel to $(1, 1, 1)$.The line $L$ passes through $(1, -1, -1)$ and is parallel to $\vec{PQ}$,so its equation is $\frac{x-1}{1} = \frac{y+1}{1} = \frac{z+1}{1} = \lambda$.Any point on $L$ is $R(\lambda+1, \lambda-1, \lambda-1)$.Since $R$ lies on the plane $x + y + z = 5$,we have $(\lambda+1) + (\lambda-1) + (\lambda-1) = 5$,which gives $3\lambda - 1 = 5$,so $3\lambda = 6$ and $\lambda = 2$.Thus,$R = (2+1, 2-1, 2-1) = (3, 1, 1)$.Finally,$QR^{2} = (3-3)^{2} + (2-1)^{2} + (3-1)^{2} = 0^{2} + 1^{2} + 2^{2} = 1 + 4 = 5$.

Let $Q$ be the mirror image of the point $P(1, 0, 1)$ with respect to the plane $S: x + y + z = 5$. If a line $L$ passing through $(1, -1, -1)$,parallel to the line $PQ$,meets the plane $S$ at $R$,then $QR^{2}$ is equal to

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