Let $\mathrm{P}$ 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 $4 x=2 y=z$ is

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{L}_1 \equiv \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-4}{5}=\frac{\mathrm{z}-2}{1}=\lambda \\ & \mathrm{P}(\lambda+2,5 \lambda+4, \lambda+2) \\ & \mathrm{L}_2 \equiv \frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{2} \\ & \mathrm{P}(2 \mu+3,3 \mu+2,2 \mu+3) \\ & \lambda+2=2 \mu+3 \quad 3 \mu+2=5 \lambda+4 \\ & \lambda=2 \mu+1 \quad 3 \mu=5 \lambda+2 \\ & 3 \mu=5(2 \mu+1)+2 \\ & 3 \mu=10 \mu+7 \quad \\ & \mu=-1 \quad \lambda=-1 \end{aligned}$
Both satisfies (P) $\begin{aligned} & \mathrm{P}(1,-1,1) \\ & \mathrm{L}_3 \equiv \frac{\mathrm{x}}{1 / 4}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{1} \\ & \mathrm{~L}_3=\frac{\mathrm{x}}{\mathrm{l}}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{4}=\mathrm{k} \end{aligned}$
Coordinates of $\mathrm{Q}(\mathrm{k}, 2 \mathrm{k}, 4 \mathrm{k})$ $\mathrm{DR}$ 's $\mathrm{PQ}=\langle\mathrm{k}-1,2 \mathrm{k}+1,4 \mathrm{k}-1>$ $\mathrm{PQ} \perp$ to $\mathrm{L}_3$ $\begin{aligned} & (\mathrm{k}-1)+2(2 \mathrm{k}+1)+4(4 \mathrm{k}-1)=0 \\ & \mathrm{k}-1+4 \mathrm{k}+2+16 \mathrm{k}-4=0 \\ & \mathrm{k}=\frac{1}{7} \\ & \mathrm{Q}\left(\frac{1}{7}, \frac{2}{7}, \frac{4}{7}\right) \\ & \mathrm{PQ}=\sqrt{\left(1-\frac{1}{7}\right)^2+\left(-1-\frac{2}{7}\right)^2+\left(1-\frac{4}{7}\right)^2} \\ & =\sqrt{\frac{36}{49}+\frac{81}{49}+\frac{9}{49}}=\frac{\sqrt{126}}{7} \\ & P Q=\frac{3 \sqrt{14}}{7} \end{aligned}$

Let $\mathrm{P}$ be the point of intersection of the lines $\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$ and…

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