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JEE Main202310 Apr 2023Morning ShiftMathematicsThree Dimensional GeometryActual

Let P be the point of intersection of the line x + 3 3 = y + 2 1 = 1 - z 2 and the plane x + y + z = 2 . If the distance of the point P from the plane 3 x - 4 y + 12 z = 32 is q , then q and 2 q are the roots of the equation

Options

  1. Ax 2 - 18 x - 72 = 0
  2. Bx 2 - 18 x + 72 = 0
  3. Cx 2 + 18 x + 72 = 0
  4. Dx 2 + 18 x - 72 = 0

Correct answer

B. x 2 - 18 x + 72 = 0

Step-by-step solution

Given, P be the point of intersection of the line x + 3 3 = y + 2 1 = 1 - z 2 and the plane x + y + z = 2 Now let, x + 3 3 = y + 2 1 = 1 - z 2 = λ Say, point A 3 λ - 3 ,   λ - 2 ,   1 - 2 λ Since point lie on the plane also, So, 3 λ - 3 + λ - 2 + 1 - 2 λ = 2 ⇒ 2 λ = 6 ⇒ λ = 3 Hence, the point will be P 6 ,   1 ,   - 5 Now finding the distance of the point P 6 ,   1 ,   - 5 from the plane 3 x - 4 y + 12 z = 32 we get, q = 18 - 4 -

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