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JEE Main202127 Aug 2021Morning ShiftMathematicsThree Dimensional GeometryActual

Equation of a plane at a distance 2 21 units from the origin, which contains the line of intersection of the planes x - y - z - 1 = 0 and 2 x + y - 3 z + 4 = 0 , is

Options

  1. A- x + 2 y + 2 z - 3 = 0
  2. B3 x - 4 z + 3 = 0
  3. C3 x - 1 y - 5 z + 2 = 0
  4. D4 x - y - 5 z + 2 = 0

Correct answer

D. 4 x - y - 5 z + 2 = 0

Step-by-step solution

Required equation of plane P 1 + λ P 2 = 0 x - y - z - 1 + λ 2 x + y - 3 z + 4 = 0 Given that its distance from origin is 2 21 Thus, 4 λ - 1 2 λ + 1 2 + λ - 1 2 + - 3 λ - 1 2 = 2 21 21 4 λ - 1 2 = 2 14 λ 2 + 8 λ + 3 336 λ 2 - 168 λ + 21 = 28 λ 2 + 16 λ + 6 308 λ 2 - 184 λ + 15 = 0 308 λ 2 - 154 λ - 30 λ + 15 = 0 2 λ - 1 154 λ - 15 = 0 λ = 1 2   &   15 154 For λ = 1 2 , required plane is 4 x -

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