Let $P Q R$ be a triangle with $R(-1,4,2)$. Suppose $M(2,1,2)$ is the mid point of $\mathrm{PQ}$. The distance of the centroid of $\triangle \mathrm{PQR}$ from the point of intersection of the lines $\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$ and $\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$ is [JEE Main 2024, 29 Jan (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: b) $\sqrt{69}$. G is centroid of $\triangle PQR$ $∴G\equiv \left(\frac{2\times 2−1}{3},\frac{2\times 1+4}{3},\frac{2\times 2+2}{3}\right)$ $G\equiv (1,2,2)$ Let $x$ is the point of intersection of lines ${l}_{1}:\frac{x−2}{0}=\frac{y}{2}=\frac{z+3}{−1}$ ${k}_{2}:\frac{x−1}{1}=\frac{y+3}{−3}=\frac{z+1}{1},\text{ then }x\equiv (2,−6,0)$ $∴$distance between $G$ and $x=\sqrt{69}$

Let \(P Q R\) be a triangle with \(R(-1,4,2)\). Suppose \(M(2,1,2)\) is the mid point of \(\mathrm{PQ}\). The distance o…

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