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The equation of a plane containing the line formed by 2 x - y + z - 3 = 0 , 3 x + y + z = 5 and at a distance of 1 6 units from the point 2,1 , - 1 is

Options

  1. Ax + y + z - 3 = 0
  2. B2 x - y - z - 3 = 0
  3. C2 x - y + z + 3 = 0
  4. D62 x + 29 y + 19 z - 105 = 0

Correct answer

D. 62 x + 29 y + 19 z - 105 = 0

Step-by-step solution

Equation of the required plane is 2 x - y + z - 3 + λ 3 x + y + z - 5 = 0 ⇒ 2 + 3 λ x + λ - 1 y + λ + 1 z - 3 - 5 λ = 0 Distance of the plane from 2,1 , - 1 = 1 6 4 + 6 λ + λ - 1 - λ - 1 - 3 - 5 λ 2 + 3 λ 2 + λ - 1 2 + λ + 1 2 = 1 6 λ - 1 11 λ 2 + 12 λ + 6 = 1 6 6 λ - 1 2 = 11 λ 2 + 12 λ + 6 5 λ 2 + 24 λ = 0 ⇒ λ = 0 , - 24 5 Hence, the equation of the plane is 62 x + 29 y + 19 z - 105 = 0

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