JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let a triangle \(PQR\) be such that \(P\) and \(Q\) lie on the line \(\dfrac{x+3}{8} = \dfrac{y-4}{2} = \dfrac{z+1}{2}\) and are at a distance of \(6\) units from \(R(1, 2, 3)\). If \((\alpha, \beta, \gamma)\) is the centroid of \(\triangle PQR\), then \(\alpha + \beta + \gamma\) is equal to :
- A \(4\)
- B \(5\)
- C \(6\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
The equation of the given line is \(\dfrac{x+3}{8} = \dfrac{y-4}{2} = \dfrac{z+1}{2}\). Let \(\dfrac{x+3}{4} = \dfrac{y-4}{1} = \dfrac{z+1}{1} = \lambda\). Any point on this line can be taken as \(S(4\lambda - 3, \lambda + 4, \lambda - 1)\). Since the points \(P\) and \(Q\) lie…
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