JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(Q\) and \(R\) be two points on the line \(\frac{ x +1}{2}=\frac{ y +2}{3}=\frac{ z -1}{2}\) at a distance \(\sqrt{26}\) from the point \(P (4,2,7)\). Then the square of the area of the triangle \(PQR\) is \(....\)
- A \(153\)
- B \(154\)
- C \(155\)
- D \(156\)
Answer & Solution
Correct Answer
(A) \(153\)
Step-by-step Solution
Detailed explanation
Let \((2 \lambda-1,3 \lambda-2,2 \lambda+1)\) be any point on the line \((2 \lambda-5)^{2}+(3 \lambda-4)^{2}+(2 \lambda-6)^{2}=26\) \(\lambda=1,3\) \(Q (1,1,3) ; R (5,7,7) ; \quad P (4,2,7)\) Area of triangle \(PQR =1 / 2|\overrightarrow{ PQ } \times \overrightarrow{ PR }|\)…
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