What is the equation of a plane that passes through the point (1, 1, 1) and is orthogonal to the planes 2x + y - 2z = 5 and 3x - 6y - 2z = 7?

4x + 2y + 15z = 31
4x - 2y + 15z = 27
-4x + 2y + 15z = 3
4x + 2y - 15z = 1

Correct answer: 4x + 2y + 15z = 31

Solution

To find the equation of the plane, we first determine the direction ratios of the normal to the required plane, which are found to be proportional to 4, 2, and 15. Using the point (1, 1, 1) that the plane passes through, we can then derive the equation of the plane as 4x + 2y + 15z = 31.

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