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JEE Main202229 Jun 2022Evening ShiftMathematicsThree Dimensional GeometryActual

Let x - 2 3 = y + 1 - 2 = z + 3 - 1 lie on the plane p x - q y + z = 5 , for some p , q ∈ R . The shortest distance of the plane from the origin is:

Options

  1. A3 109
  2. B5 142
  3. C5 71
  4. D1 142

Correct answer

B. 5 142

Step-by-step solution

Given, line x - 2 3 = y + 1 - 2 = z + 3 - 1 So, the point 2 ,   - 1 ,   - 3 will satisfy the given plane p x - q y + z = 5 So, 2 p + q - 3 = 5 or 2 p + q = 8 ... (i) Also given line is perpendicular to normal plane so 3 p + 2 q - 1 = 0 ... (ii) ⇒ p = 15 ,   q = - 22 So,equation of plane 15 x - 22 y + z - 5 = 0 Now its distance from origin = 6 710 = 5 142

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