Question:
The plane containing the line \(\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z-1}{3}\) and also containing its projection on the plane \(2x + 3y - z = 5\), contains which one of the following points ?
The plane containing the line \(\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z-1}{3}\) and also containing its projection on the plane \(2x + 3y - z = 5\), contains which one of the following points ?
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The required plane contains the line direction and the normal of the given plane (it is perpendicular to the given plane), so its normal is the cross product of these two vectors.
Updated On: Nov 5, 2026
- \((-2, 2, 2)\)
- \((2, 2, 0)\)
- \((2, 0, -2)\)
- \((0, -2, 2)\)
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The Correct Option is C
Solution and Explanation
Direction of the line: \(\vec{d} = (2, -1, 3)\); it passes through \((3, -2, 1)\).
The plane containing the line and its projection on \(2x+3y-z=5\) is perpendicular to that plane, so it contains the normal \(\vec{n} = (2, 3, -1)\).
Normal of the required plane: \(\vec{d}\times\vec{n} = (-1\cdot(-1) - 3\cdot 3,\; 3\cdot 2 - 2\cdot(-1),\; 2\cdot 3 - (-1)\cdot 2) = (-8, 8, 8)\), i.e. direction ratios \((1, -1, -1)\).
Plane through \((3,-2,1)\): \((x-3) - (y+2) - (z-1) = 0 \Rightarrow x - y - z = 4\).
Check options: \((-2,2,2)\) gives \(-6\); \((2,2,0)\) gives \(0\); \((2,0,-2)\) gives \(2 - 0 + 2 = 4\) (satisfied); \((0,-2,2)\) gives \(0+2-2 = 0\).
Answer: (3) \((2, 0, -2)\)
The plane containing the line and its projection on \(2x+3y-z=5\) is perpendicular to that plane, so it contains the normal \(\vec{n} = (2, 3, -1)\).
Normal of the required plane: \(\vec{d}\times\vec{n} = (-1\cdot(-1) - 3\cdot 3,\; 3\cdot 2 - 2\cdot(-1),\; 2\cdot 3 - (-1)\cdot 2) = (-8, 8, 8)\), i.e. direction ratios \((1, -1, -1)\).
Plane through \((3,-2,1)\): \((x-3) - (y+2) - (z-1) = 0 \Rightarrow x - y - z = 4\).
Check options: \((-2,2,2)\) gives \(-6\); \((2,2,0)\) gives \(0\); \((2,0,-2)\) gives \(2 - 0 + 2 = 4\) (satisfied); \((0,-2,2)\) gives \(0+2-2 = 0\).
Answer: (3) \((2, 0, -2)\)
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