Question:
If the point \( (2, \alpha, \beta) \) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane \( 2x - 5y = 15 \), then \( 2\alpha - 3\beta \) is equal to
If the point \( (2, \alpha, \beta) \) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane \( 2x - 5y = 15 \), then \( 2\alpha - 3\beta \) is equal to
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The required plane contains the vector joining the two points and the normal of the given plane; take their cross product as the normal.
Updated On: Nov 5, 2026
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The Correct Option is B
Solution and Explanation
Vector along the plane: \( \vec{AB} = (4, -4, 4) \parallel (1, -1, 1) \). Normal of given plane: \( (2, -5, 0) \).
Normal of required plane \( = (1,-1,1) \times (2,-5,0) = (0+5,\ 2-0,\ -5+2) = (5, 2, -3) \).
Plane through (3, 4, 2): \( 5(x-3) + 2(y-4) - 3(z-2) = 0 \Rightarrow 5x + 2y - 3z = 17 \).
Point \( (2,\alpha,\beta) \): \( 10 + 2\alpha - 3\beta = 17 \Rightarrow 2\alpha - 3\beta = 7 \).
Answer: (2) 7
Normal of required plane \( = (1,-1,1) \times (2,-5,0) = (0+5,\ 2-0,\ -5+2) = (5, 2, -3) \).
Plane through (3, 4, 2): \( 5(x-3) + 2(y-4) - 3(z-2) = 0 \Rightarrow 5x + 2y - 3z = 17 \).
Point \( (2,\alpha,\beta) \): \( 10 + 2\alpha - 3\beta = 17 \Rightarrow 2\alpha - 3\beta = 7 \).
Answer: (2) 7
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