Question:
Let P be a plane passing through the points (1, 0, 1), (1, -2, 1) and (0, 1, -2). Let a vector \(\vec{a} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}\) be such that \(\vec{a}\) is parallel to the plane P, perpendicular to \((\hat{i} + 2\hat{j} + 3\hat{k})\) and \(\vec{a}\cdot(\hat{i} + \hat{j} + 2\hat{k}) = 2\), then \((\alpha - \beta + \gamma)^2\) equals ____________.
Let P be a plane passing through the points (1, 0, 1), (1, -2, 1) and (0, 1, -2). Let a vector \(\vec{a} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}\) be such that \(\vec{a}\) is parallel to the plane P, perpendicular to \((\hat{i} + 2\hat{j} + 3\hat{k})\) and \(\vec{a}\cdot(\hat{i} + \hat{j} + 2\hat{k}) = 2\), then \((\alpha - \beta + \gamma)^2\) equals ____________.
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Find the normal of the plane, then impose the three conditions on \((\alpha, \beta, \gamma)\).
Updated On: Nov 4, 2026
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Correct Answer: 81
Solution and Explanation
Vectors in the plane: \((0, -2, 0)\) and \((-1, 1, -3)\). Normal \(= (0,-2,0)\times(-1,1,-3) = (6, 0, -2) \parallel (3, 0, -1)\).
\(\vec{a}\) parallel to the plane: \(3\alpha - \gamma = 0\), so \(\gamma = 3\alpha\).
\(\vec{a}\perp(\hat{i}+2\hat{j}+3\hat{k})\): \(\alpha + 2\beta + 3\gamma = 0\), so \(10\alpha + 2\beta = 0\), \(\beta = -5\alpha\).
\(\vec{a}\cdot(\hat{i}+\hat{j}+2\hat{k}) = \alpha + \beta + 2\gamma = \alpha - 5\alpha + 6\alpha = 2\alpha = 2\), so \(\alpha = 1\), \(\beta = -5\), \(\gamma = 3\).
\((\alpha - \beta + \gamma)^2 = (1 + 5 + 3)^2 = 81\).
Answer: 81
\(\vec{a}\) parallel to the plane: \(3\alpha - \gamma = 0\), so \(\gamma = 3\alpha\).
\(\vec{a}\perp(\hat{i}+2\hat{j}+3\hat{k})\): \(\alpha + 2\beta + 3\gamma = 0\), so \(10\alpha + 2\beta = 0\), \(\beta = -5\alpha\).
\(\vec{a}\cdot(\hat{i}+\hat{j}+2\hat{k}) = \alpha + \beta + 2\gamma = \alpha - 5\alpha + 6\alpha = 2\alpha = 2\), so \(\alpha = 1\), \(\beta = -5\), \(\gamma = 3\).
\((\alpha - \beta + \gamma)^2 = (1 + 5 + 3)^2 = 81\).
Answer: 81
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