Question:

The plane through the intersection of the planes $x + y + z = 1$ and $2x + 3y - z + 4 = 0$ and parallel to y-axis also passes through the point :

Updated On: Aug 21, 2024
  • (-3, 0, -1)
  • (3, 3, -1)
  • (3, 2, 1)
  • (-3, 1, 1)
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The Correct Option is C

Solution and Explanation

Equation of plane $(x + y + z - 1) + \lambda (2x + 3y - z + 4) = 0$ $\Rightarrow \; (1 + 2 \lambda)x + (1 + 3 \lambda )y + (1 - \lambda)z - 1 +4\lambda = 0$ dr's of normal of the plane are $1 + 2\lambda, 1+ 3 \lambda, 1 - \lambda$ Since plane is parallel to y - axis, $1 + 3\lambda = 0$ $\Rightarrow \; \lambda = -1/3$ So the equation of plane is $x + 4z - 7 = 0$ Point $(3, 2, 1)$ satisfies this equation
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