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JEE Main202311 Apr 2023Evening ShiftMathematicsThree Dimensional GeometryActual

Let the line passing through the points P 2 , - 1 , 2 and Q 5 , 3 , 4 meet the plane x - y + z = 4 at the point R . Then the distance of the point R from the plane x + 2 y + 3 z + 2 = 0 measured parallel to the line x - 7 2 = y + 3 2 = z - 2 1 is

Options

  1. A61
  2. B189
  3. C31
  4. D3

Correct answer

D. 3

Step-by-step solution

Given, The line passing through the points P 2 , - 1 , 2 and Q 5 , 3 , 4 , So, equation of line P Q will be, x - 2 3 = y + 1 4 = z - 2 2 = λ Now let point R be 3 λ + 2 , 4 λ - 1 , 2 λ + 2 Given, R lies on plane x - y + z = 4 ∴   3 λ + 2 - 4 λ + 1 + 2 λ + 2 = 4 ⇒ λ = - 1 ∴   R - 1 , - 5 , 0 Now let line S R be : x + 1 2 = y + 5 2 = z 1 = k as it is parallel to x - 7 2 = y + 3 2 = z - 2 1 and passing through - 1 , - 5 , 0 So, the point S : 2 k - 1 ,

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