Question:
If the plane \(y = \alpha x - \beta z + \gamma\) passing through the point \((1, -1, 3)\) is perpendicular to each of the planes \(2x + y + z = 1\) and \(3x - 2y + 2z = 0\), then \(\alpha + \beta + \gamma\) is equal to :
If the plane \(y = \alpha x - \beta z + \gamma\) passing through the point \((1, -1, 3)\) is perpendicular to each of the planes \(2x + y + z = 1\) and \(3x - 2y + 2z = 0\), then \(\alpha + \beta + \gamma\) is equal to :
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Write the plane as \(\alpha x - y - \beta z + \gamma = 0\); its normal must be perpendicular to the normals of the two given planes.
Updated On: Nov 12, 2026
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The Correct Option is D
Solution and Explanation
The plane is \(\alpha x - y - \beta z + \gamma = 0\) with normal \((\alpha, -1, -\beta)\).
Perpendicular to \(2x + y + z = 1\): \(2\alpha - 1 - \beta = 0\).
Perpendicular to \(3x - 2y + 2z = 0\): \(3\alpha + 2 - 2\beta = 0\).
From the first, \(\beta = 2\alpha - 1\). Substituting: \(3\alpha + 2 - 4\alpha + 2 = 0\), so \(\alpha = 4\) and \(\beta = 7\).
The plane passes through (1, -1, 3): \(-1 = \alpha(1) - \beta(3) + \gamma = 4 - 21 + \gamma\), so \(\gamma = 16\).
\(\alpha + \beta + \gamma = 4 + 7 + 16 = 27\).
Answer: (4) 27
Perpendicular to \(2x + y + z = 1\): \(2\alpha - 1 - \beta = 0\).
Perpendicular to \(3x - 2y + 2z = 0\): \(3\alpha + 2 - 2\beta = 0\).
From the first, \(\beta = 2\alpha - 1\). Substituting: \(3\alpha + 2 - 4\alpha + 2 = 0\), so \(\alpha = 4\) and \(\beta = 7\).
The plane passes through (1, -1, 3): \(-1 = \alpha(1) - \beta(3) + \gamma = 4 - 21 + \gamma\), so \(\gamma = 16\).
\(\alpha + \beta + \gamma = 4 + 7 + 16 = 27\).
Answer: (4) 27
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