Let the image of the point (1,0,7) in the lin | vedclass

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(C) Let the line be $L_1: \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda$. Any point $M$ on the line is $(\lambda, 1+2\lambda, 2+3\lambda)$.Let $P =

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$(3,4,3-2 \sqrt{2})$

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(C) Let the line be $L_1: \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda$. Any point $M$ on the line is $(\lambda, 1+2\lambda, 2+3\lambda)$.Let $P = (1,0,7)$. The vector $\overrightarrow{PM} = (\lambda-1, 1+2\lambda, 3\lambda-5)$.Since $\overrightarrow{PM}$ is perpendicular to the line $L_1$ with direction vector $\vec{b} = (1,2,3)$,we have $\overrightarrow{PM} \cdot \vec{b} = 0$.$(\lambda-1)(1) + (1+2\lambda)(2) + (3\lambda-5)(3) = 0 \Rightarrow \lambda-1+2+4\lambda+9\lambda-15 = 0 \Rightarrow 14\lambda = 14 \Rightarrow \lambda = 1$.Thus,$M = (1, 3, 5)$.Since $M$ is the midpoint of $PQ$,where $Q = (\alpha, \beta, \gamma)$,we have $M = \frac{P+Q}{2} \Rightarrow Q = 2M - P = 2(1,3,5) - (1,0,7) = (2-1, 6-0, 10-7) = (1,6,3)$.So,$(\alpha, \beta, \gamma) = (1,6,3)$.Let the direction cosines of the required line be $(l, m, n)$. Given $m = \cos(\frac{2\pi}{3}) = -\frac{1}{2}$ and $n = \cos(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}}$.Since $l^2+m^2+n^2=1$,$l^2 + (-\frac{1}{2})^2 + (-\frac{1}{\sqrt{2}})^2 = 1 \Rightarrow l^2 + \frac{1}{4} + \frac{1}{2} = 1 \Rightarrow l^2 = \frac{1}{4}$.Since the line makes an acute angle with the $x$-axis,$l = \frac{1}{2}$.The line passes through $(1,6,3)$ with direction vector $(\frac{1}{2}, -\frac{1}{2}, -\frac{1}{\sqrt{2}})$,or $(1, -1, -\sqrt{2})$.The equation is $\frac{x-1}{1} = \frac{y-6}{-1} = \frac{z-3}{-\sqrt{2}} = \mu$.For $\mu = 2$,$x = 3, y = 4, z = 3-2\sqrt{2}$. This matches option $C$.

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