Question:
Let \(L\) be a line obtained from the intersection of two planes \(x + 2y + z = 6\) and \(y + 2z = 4\). If point \(P(\alpha, \beta, \gamma)\) is the foot of perpendicular from \((3, 2, 1)\) on \(L\), then the value of \(21(\alpha + \beta + \gamma)\) equals :
Let \(L\) be a line obtained from the intersection of two planes \(x + 2y + z = 6\) and \(y + 2z = 4\). If point \(P(\alpha, \beta, \gamma)\) is the foot of perpendicular from \((3, 2, 1)\) on \(L\), then the value of \(21(\alpha + \beta + \gamma)\) equals :
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Direction of \(L\) is the cross product of the plane normals; write a general point on \(L\) and impose perpendicularity of the joining vector to the direction.
Updated On: Nov 4, 2026
- \(102\)
- \(142\)
- \(136\)
- \(68\)
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The Correct Option is A
Solution and Explanation
Direction of \(L\): \((1,2,1)\times(0,1,2) = (3, -2, 1)\).
A point on \(L\): put \(z = 0\): \(y = 4\), \(x = -2\), so \((-2, 4, 0)\).
Points of \(L\): \((-2 + 3t,\ 4 - 2t,\ t)\). Vector from \(A(3,2,1)\) to it: \((-5 + 3t,\ 2 - 2t,\ t - 1)\).
Perpendicular to \((3,-2,1)\): \(3(-5+3t) - 2(2-2t) + (t-1) = 14t - 20 = 0\), \(t = \frac{10}{7}\).
\(P = \left(\frac{16}{7}, \frac{8}{7}, \frac{10}{7}\right)\), so \(\alpha + \beta + \gamma = \frac{34}{7}\) and \(21(\alpha+\beta+\gamma) = 102\).
Answer: (1) \(102\)
A point on \(L\): put \(z = 0\): \(y = 4\), \(x = -2\), so \((-2, 4, 0)\).
Points of \(L\): \((-2 + 3t,\ 4 - 2t,\ t)\). Vector from \(A(3,2,1)\) to it: \((-5 + 3t,\ 2 - 2t,\ t - 1)\).
Perpendicular to \((3,-2,1)\): \(3(-5+3t) - 2(2-2t) + (t-1) = 14t - 20 = 0\), \(t = \frac{10}{7}\).
\(P = \left(\frac{16}{7}, \frac{8}{7}, \frac{10}{7}\right)\), so \(\alpha + \beta + \gamma = \frac{34}{7}\) and \(21(\alpha+\beta+\gamma) = 102\).
Answer: (1) \(102\)
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