Let Q(a,b,c) be the image of the point P(3,2, | vedclass

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(C) Let the line be $L_1: \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} = r$. Any point $N$ on $L_1$ is $(r+1, 2r, r+1)$.The direction ratios of $PN$ are $(

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$7$

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(C) Let the line be $L_1: \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} = r$. Any point $N$ on $L_1$ is $(r+1, 2r, r+1)$.The direction ratios of $PN$ are $(r+1-3, 2r-2, r+1-1) = (r-2, 2r-2, r)$.Since $PN$ is perpendicular to $L_1$ (direction vector $\vec{v_1} = \langle 1, 2, 1 \rangle$),we have $1(r-2) + 2(2r-2) + 1(r) = 0$.$r-2 + 4r-4 + r = 0 \Rightarrow 6r = 6 \Rightarrow r = 1$.Thus,$N = (1+1, 2(1), 1+1) = (2, 2, 2)$.Since $N$ is the midpoint of $PQ$,let $Q = (x_q, y_q, z_q)$. Then $\frac{x_q+3}{2} = 2, \frac{y_q+2}{2} = 2, \frac{z_q+1}{2} = 2$.$x_q = 1, y_q = 2, z_q = 3$. So $Q = (1, 2, 3)$.Now,find the distance of $Q(1, 2, 3)$ from the line $L_2: \frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2} = k$.Any point $M$ on $L_2$ is $(3k+9, 2k+9, -2k+5)$.The vector $\vec{QM} = \langle 3k+8, 2k+7, -2k+2 \rangle$. The direction of $L_2$ is $\vec{v_2} = \langle 3, 2, -2 \rangle$.Since $\vec{QM} \perp \vec{v_2}$,$3(3k+8) + 2(2k+7) - 2(-2k+2) = 0$.$9k+24 + 4k+14 + 4k-4 = 0 \Rightarrow 17k + 34 = 0 \Rightarrow k = -2$.Substituting $k=-2$,$M = (3(-2)+9, 2(-2)+9, -2(-2)+5) = (3, 5, 9)$.The distance $QM = \sqrt{(3-1)^2 + (5-2)^2 + (9-3)^2} = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$.

Let $Q(a,b,c)$ be the image of the point $P(3,2,1)$ in the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}.$ Then the distance of $Q$ from the line $\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$ is

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