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The equation of the plane passing through the point (1, 2, −3) and perpendicular to the planes 3x + y − 2z = 5 and 2x − 5y − z = 7, is (1) 3x − 10y − 2z + 11 = 0 (2) 6x − 5y − 2z − 0 = 0 (3) 11x + y + 17z + 38 = 0 (4) 6x − 5y + 2z + 10 = 0
- 3x − 10y − 2z + 11 = 0
- 6x − 5y − 2z − 0 = 0
- 11x + y + 17z + 38 = 0
- 6x − 5y + 2z + 10 = 0
Correct answer: 6x − 5y + 2z + 10 = 0
Solution
The correct option represents a plane that is perpendicular to the normal vectors of the given planes, which can be determined by taking the cross product of their normal vectors. The resulting normal vector, along with the point (1, 2, -3), is used to derive the equation of the desired plane.
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