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JEE Main202124 Feb 2021Morning ShiftMathematicsThree Dimensional GeometryActual

The equation of the plane passing through the point 1 , 2 , - 3 and perpendicular to the planes 3 x + y - 2 z = 5 and 2 x - 5 y - z = 7 , is

Options

  1. A11 x + y + 17 z + 38 = 0
  2. B3 x - 10 y - 2 z + 11 = 0
  3. C6 x - 5 y + 2 z + 10 = 0
  4. D6 x - 5 y - 2 z - 2 = 0

Correct answer

A. 11 x + y + 17 z + 38 = 0

Step-by-step solution

Normal vector of the plane will be i ^ j ^ k ^ 3 1 - 2 2 - 5 - 1 = - 11 i ^ - j ^ - 17 k ^ So, the direction ratios of normal to the required plane are 11 , 1 , 17 and plane passes through 1 ,   2 , - 3 . So, the equation of plane is 11 x - 1 + 1 y - 2 + 17 z + 3 = 0 ⇒ 11 x + y + 17 z + 38 = 0

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