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(B) Let $P$ be a point on the line $x = y + 2 = z = t$. Then $P = (t, t - 2, t)$.Let $Q$ be a point on the line $x + 2 = 2y = 2z = 2s$. Then $Q = (2s

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$65$

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(B) Let $P$ be a point on the line $x = y + 2 = z = t$. Then $P = (t, t - 2, t)$.Let $Q$ be a point on the line $x + 2 = 2y = 2z = 2s$. Then $Q = (2s - 2, s, s)$.The direction ratios of the line $PQ$ are given as $(2, 1, 2)$.Thus,the direction ratios of the vector $\vec{PQ}$ are $(2s - 2 - t, s - (t - 2), s - t) = (2s - t - 2, s - t + 2, s - t)$.Since the line $PQ$ has direction ratios $(2, 1, 2)$,we have:$\frac{2s - t - 2}{2} = \frac{s - t + 2}{1} = \frac{s - t}{2} = k$ (say).From $\frac{s - t + 2}{1} = \frac{s - t}{2}$,we get $2s - 2t + 4 = s - t$,so $s - t = -4$.Substituting $s - t = -4$ into the ratios:$\frac{2s - t - 2}{2} = \frac{-4 + 2}{1} = -2$,so $2s - t - 2 = -4$,which means $2s - t = -2$.Solving $s - t = -4$ and $2s - t = -2$,we get $s = 2$ and $t = 6$.Thus,$P = (6, 4, 6)$ and $Q = (2, 2, 2)$.The equation of line $PQ$ is $\frac{x - 2}{2} = \frac{y - 2}{1} = \frac{z - 2}{2} = \lambda$.Any point $F$ on $PQ$ is $(2\lambda + 2, \lambda + 2, 2\lambda + 2)$.Let $A = (1, 2, 12)$. The vector $\vec{AF} = (2\lambda + 1, \lambda, 2\lambda - 10)$.Since $AF \perp PQ$,$\vec{AF} \cdot (2, 1, 2) = 0$.$2(2\lambda + 1) + 1(\lambda) + 2(2\lambda - 10) = 0$.$4\lambda + 2 + \lambda + 4\lambda - 20 = 0 \Rightarrow 9\lambda = 18 \Rightarrow \lambda = 2$.So $F = (6, 4, 6)$.The length $l = AF = \sqrt{(6 - 1)^2 + (4 - 2)^2 + (6 - 12)^2} = \sqrt{5^2 + 2^2 + (-6)^2} = \sqrt{25 + 4 + 36} = \sqrt{65}$.Therefore,$l^2 = 65$.

$A$ line with direction ratios $2, 1, 2$ meets the lines $x = y + 2 = z$ and $x + 2 = 2y = 2z$ at points $P$ and $Q$ respectively. If the length of the perpendicular from the point $(1, 2, 12)$ to the line $PQ$ is $l$,then $l^2$ is

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