Let L_1: frac{x-1}{2}=frac{y-2}{3}=frac{z-3}{ | vedclass

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(D) Let $P(2\lambda+1, 3\lambda+2, 4\lambda+3)$ be a point on $L_1$ and $Q(3\mu+2, 4\mu+4, 5\mu+5)$ be a point on $L_2$.The direction ratios of $PQ$ a

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$\left(\frac{14}{3},-3, \frac{22}{3}\right)$

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(D) Let $P(2\lambda+1, 3\lambda+2, 4\lambda+3)$ be a point on $L_1$ and $Q(3\mu+2, 4\mu+4, 5\mu+5)$ be a point on $L_2$.The direction ratios of $PQ$ are $(3\mu-2\lambda+1, 4\mu-3\lambda+2, 5\mu-4\lambda+2)$.Since $PQ$ is the shortest distance line,$PQ \perp L_1$ and $PQ \perp L_2$.For $PQ \perp L_1$: $2(3\mu-2\lambda+1) + 3(4\mu-3\lambda+2) + 4(5\mu-4\lambda+2) = 0 \Rightarrow 38\mu - 29\lambda + 16 = 0$.For $PQ \perp L_2$: $3(3\mu-2\lambda+1) + 4(4\mu-3\lambda+2) + 5(5\mu-4\lambda+2) = 0 \Rightarrow 50\mu - 38\lambda + 21 = 0$.Solving these equations,we get $\lambda = \frac{1}{3}$ and $\mu = -\frac{1}{6}$.Substituting these values,$P = \left(\frac{5}{3}, 3, \frac{13}{3}\right)$ and $Q = \left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right)$.The equation of the line $PQ$ passing through $P$ with direction ratios proportional to $\vec{PQ} = Q-P = \left(-\frac{1}{6}, \frac{1}{3}, -\frac{1}{6}\right)$ (or $(1, -2, 1)$) is $\frac{x-5/3}{1} = \frac{y-3}{-2} = \frac{z-13/3}{1}$.Checking the options,the point $\left(\frac{14}{3}, -3, \frac{22}{3}\right)$ satisfies this equation: $\frac{14/3 - 5/3}{1} = 3$,$\frac{-3-3}{-2} = 3$,$\frac{22/3 - 13/3}{1} = 3$.

Let $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$?

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