JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(P Q R\) be a triangle with \(R(-1,4,2)\). Suppose \(M(2,1,2)\) is the mid point of \(PQ\). The distance of the centroid of \(\triangle \mathrm{PQR}\) from the point of intersection of the line \(\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}\) and \(\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}\) is
- A \(69\)
- B \(9\)
- C \(\sqrt{69}\)
- D \(\sqrt{99}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{69}\)
Step-by-step Solution
Detailed explanation
Centroid \(G\) divides \(M R\) in \(1: 2\) \(\mathrm{G}(1,2,2)\) Point of intersection \(A\) of given lines is \((2,-6,0)\) \(\mathrm{AG}=\sqrt{69}\)
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