JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the plane containing the line of intersection of the planes \(P 1: x+(\lambda+4) y+z=1\) and \(P2:\) \(2 x+y+z=2\) pass through the points \((0,1,0)\) and \((1,0,1)\). Then the distance of the point \((2 \lambda, \lambda,-\lambda)\) from the plane \(P 2\) is
- A \(5 \sqrt{6}\)
- B \(4 \sqrt{6}\)
- C \(2 \sqrt{6}\)
- D \(3 \sqrt{6}\)
Answer & Solution
Correct Answer
(D) \(3 \sqrt{6}\)
Step-by-step Solution
Detailed explanation
Equation of plane passing through point of intersection of \(P 1\) and \(P 2\) \(P = P 1+ kP 2\) \(( x +(\lambda+4) y + z -1)+ k (2 x + y + z -2)=0\) Passing through \((0,1,0)\) and \((1,0,1)\) \((\lambda+4-1)+ k (1-2)=0\) \((\lambda+3)- k =0\) Also passing \((1,0,1)\)…
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