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Question

Find the equation of the plane passing through the intersection of the planes 2x + 3y − z + 1 = 0 and x + y − 2z + 3 = 0 and perpendicular to the plane 3x − y − 2z − 4 = 0.

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Solution

The equation of the plane passing through the line of intersection of the given planes is2x + 3y - z + 1 + λ x + y - 2z + 3 = 0 2 + λx + 3 + λy + -1 - 2λz + 1 + 3λ = 0... 1This plane is perpendicular to 3x - y - 2z - 4 = 0. So,3 2 + λ - 3 + λ - 2 -1 - 2λ = 0 (Because a1a2 + b1b2 + c1c2 = 0)6 + 3λ - 3 - λ + 2 + 4λ = 06λ + 5 = 0λ = -5 6 Substituting this in (1), we get2 - 56x + 3 - 56y + -1 - 2 -56z + 1 + 3 -56 = 07X + 13Y + 4z - 9 = 0

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