If the image of the point P(1, -2, 1) with re | vedclass

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(D) The line passes through $B(1, 1, 2)$ and $C(2, 2, 1)$. The direction ratios of the line are $(2-1, 2-1, 1-2) = (1, 1, -1)$.The equation of the lin

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$26$

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(D) The line passes through $B(1, 1, 2)$ and $C(2, 2, 1)$. The direction ratios of the line are $(2-1, 2-1, 1-2) = (1, 1, -1)$.The equation of the line is $\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-2}{-1} = k$.Any point $Q$ on the line is $(k+1, k+1, -k+2)$.The vector $\vec{PQ} = (k+1-1, k+1-(-2), -k+2-1) = (k, k+3, 1-k)$.Since $\vec{PQ}$ is perpendicular to the line, the dot product of $\vec{PQ}$ and the direction vector $(1, 1, -1)$ is zero:$1(k) + 1(k+3) - 1(1-k) = 0$$k + k + 3 - 1 + k = 0 \Rightarrow 3k + 2 = 0 \Rightarrow k = -\frac{2}{3}$.Substituting $k$ to find $Q$: $x = -\frac{2}{3} + 1 = \frac{1}{3}$, $y = -\frac{2}{3} + 1 = \frac{1}{3}$, $z = -(-\frac{2}{3}) + 2 = \frac{8}{3}$.$Q$ is the midpoint of $PR$, where $R = (l, m, n)$:$\frac{1+l}{2} = \frac{1}{3} \Rightarrow l = -\frac{1}{3}$.$\frac{-2+m}{2} = \frac{1}{3} \Rightarrow m = \frac{8}{3}$.$\frac{1+n}{2} = \frac{8}{3} \Rightarrow n = \frac{13}{3}$.Therefore, $l^2 + m^2 + n^2 = (-\frac{1}{3})^2 + (\frac{8}{3})^2 + (\frac{13}{3})^2 = \frac{1+64+169}{9} = \frac{234}{9} = 26$.

If the image of the point $P(1, -2, 1)$ with respect to the line passing through the points $B(1, 1, 2)$ and $C(2, 2, 1)$ is $R(l, m, n)$, then $l^2 + m^2 + n^2 =$

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