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Question: Answered & Verified by Expert
Let L be the line through the intersection of the planes 3x-y+2z+1=0 and 3x-2y+z=3. Then, the equation of the plane passing through 2,1,4 and perpendicular to the line L is
MathematicsThree Dimensional GeometryJEE Main
Options:
  • A x+y-z=2
  • B x+y-z+1=0
  • C x+y+z-7=0
  • D 2x-3y+4z=17
Solution:
2484 Upvotes Verified Answer
The correct answer is: x+y-z+1=0
A vector parallel to the line of intersection is i^j^k^3-123-21
=i^3-j^-3+k^-3
=3i^+3j^-3k^
Also, the line of intersection is perpendicular to the required plane.
Hence, equation of the plane is 1x-2+1y-1-1z-4=0
x+y-z+1=0

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