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} = \fr

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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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drs of $PN = < r - 2,2r - 2,r >$$${1.(r - 2) + 2(2r - 2) + 1.(r) = 0 }{6r = 6 \Rightarrow r = 1 }{	herefore N \equiv (2,2,2) }{\Rightarrow Q \equiv (1,2,3) }$$$${AQ = \sqrt{64 + 49 + 4} = \sqrt{117} }{AM = \left| \frac{24 + 14 - 4}{\sqrt{9 + 4 + 4}} ight| = \frac{34}{\sqrt{17}} = 2\sqrt{17} }{	herefore QM = \sqrt{117 - 68} = \sqrt{49} = 7}$$

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