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Question

Find the equation of the plane which contains the line of intersection of the planes x + 2y + 3z − 4 = 0 and 2x + y − z + 5 = 0 and which is perpendicular to the plane 5x + 3y − 6z + 8 = 0.

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Solution

The equation of the plane passing through the line of intersection of the given planes isx + 2y + 3z - 4 + λ 2x + y - z + 5 = 0 1 + 2λx + 2 + λy + 3 - λz - 4 + 5λ = 0... 1This plane is perpendicular to 5x + 3y - 6z + 8 = 0. So,5 1 + 2λ + 32 + λ - 6 3 - λ = 0 (Because a1a2+b1b2+c1c2=0)5 + 10λ + 6 + 3λ - 18 + 6λ = 019λ - 7 = 0λ = 719Substituting this in (1), we get1 + 2 719x +2 + 719y + 3 - 719z - 4 + 5 719 = 033x + 45y + 50z - 41=0

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