JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(P\) and \(Q\) be the points on the line \(\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}\) which are at a distance of \(6\) units from the point \(R(1,2,3)\). If the centroid of the triangle \(PQR\) is \((\alpha, \beta, \gamma)\), then \(\alpha^2+\beta^2+\gamma^2\) is :
- A \(26\)
- B \(36\)
- C \(18\)
- D \(24\)
Answer & Solution
Correct Answer
(C) \(18\)
Step-by-step Solution
Detailed explanation
\( \mathrm{P}(8 \lambda-3,2 \lambda+4,2 \lambda-1) \) \( \mathrm{PR}=6 \) \( (8 \lambda-4)^2+(2 \lambda+2)^2+(2 \lambda-4)^2=36 \) \( \lambda=0,1 \) \( \text { Hence } \mathrm{P}(-3,4,-1) \& \mathrm{Q}(5,6,1) \)…
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