Let P be the point of intersection of the lin | vedclass

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(C) Let the first line be $L_1: \frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}=\lambda$. Any point on $L_1$ is $P(\lambda+2, 5\lambda+4, \lambda+2)$.Let th

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$\frac{3 \sqrt{14}}{7}$

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(C) Let the first line be $L_1: \frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}=\lambda$. Any point on $L_1$ is $P(\lambda+2, 5\lambda+4, \lambda+2)$.Let the second line be $L_2: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}=\mu$. Any point on $L_2$ is $P(2\mu+3, 3\mu+2, 2\mu+3)$.For intersection,$\lambda+2 = 2\mu+3$ and $5\lambda+4 = 3\mu+2$.From the first equation,$\lambda = 2\mu+1$. Substituting into the second: $5(2\mu+1)+4 = 3\mu+2 \implies 10\mu+9 = 3\mu+2 \implies 7\mu = -7 \implies \mu = -1$.Then $\lambda = 2(-1)+1 = -1$. The point of intersection $P$ is $(1, -1, 1)$.The line $L_3$ is $4x=2y=z$,which can be written as $\frac{x}{1/4} = \frac{y}{1/2} = \frac{z}{1}$ or $\frac{x}{1} = \frac{y}{2} = \frac{z}{4} = k$.Any point $Q$ on $L_3$ is $(k, 2k, 4k)$. The vector $\vec{PQ} = (k-1, 2k+1, 4k-1)$.Since $\vec{PQ}$ is perpendicular to the direction vector of $L_3$,which is $\vec{v} = (1, 2, 4)$,we have $\vec{PQ} \cdot \vec{v} = 0$.$(k-1)(1) + (2k+1)(2) + (4k-1)(4) = 0 \implies k-1 + 4k+2 + 16k-4 = 0 \implies 21k - 3 = 0 \implies k = \frac{1}{7}$.The point $Q$ is $(\frac{1}{7}, \frac{2}{7}, \frac{4}{7})$.The distance $PQ = \sqrt{(1-\frac{1}{7})^2 + (-1-\frac{2}{7})^2 + (1-\frac{4}{7})^2} = \sqrt{(\frac{6}{7})^2 + (-\frac{9}{7})^2 + (\frac{3}{7})^2} = \sqrt{\frac{36+81+9}{49}} = \sqrt{\frac{126}{49}} = \frac{\sqrt{9 	imes 14}}{7} = \frac{3\sqrt{14}}{7}$.

Let $P$ be the point of intersection of the lines $\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$ and $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$. Then,the shortest distance of $P$ from the line $4x=2y=z$ is

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