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JEE Main2017MathematicsThree Dimensional GeometryActual

The distance of the point 1 , 3 , - 7 from the plane passing through the point 1 , - 1 , - 1 , having normal perpendicular to both the lines x - 1 1 = y + 2 - 2 = z - 4 3 and x - 2 2 = y + 1 - 1 = z + 7 - 1 , is:

Options

  1. A20 74
  2. B10 83
  3. C5 83
  4. D10 74

Correct answer

B. 10 83

Step-by-step solution

Let l ,   m ,   n be the direction cosines of the line normal to the plane, then l - 2 m + 3 n = 0 and 2 l - m - n = 0 ⇒ l 2 + 3 = m 6 + 1 = n - 1 + 4 = λ ⇒ l = 5 λ ,   m = 7 λ ,   n = 3 λ ∴ The equation of the plane is 5 x + 7 y + 3 y + d = 0 ∵   it passes through   1 , - 1 , - 1 ⇒ 5 - 7 - 3 + d = 0 ⇒ d = 5 Hence, the equation of the plane is 5 x + 7 y + 3 y + 5 = 0 Now, P Q = 5 + 21 - 21 + 5 5 2 + 7 2 + 3 2 ⇒ P Q = 10 25 +

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