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JEE Main202228 Jun 2022Evening ShiftMathematicsThree Dimensional GeometryActual

Let the plane a x + b y + c z = d pass through 2 , 3 , - 5 and is perpendicular to the planes 2 x + y - 5 z = 10 and 3 x + 5 y - 7 z = 12 If a , b , c , d are integers d > 0 and g c d a , b , c , d = 1 then the value of a + 7 b + c + 20 d is equal to

Options

  1. A18
  2. B20
  3. C24
  4. D22

Correct answer

D. 22

Step-by-step solution

We know that D.R of normal of plane will be given by cross product of given two planes, as it is perpendicular to both given planes, so i ^ j ^ k ^ 2 1 - 5 3 5 - 7 = 18 i ^ - j ^ + 7 k ^ So, equation of plane is, 18 x - y + 7 z = d And it passes through 2 , 3 , - 5 So, 36 - 3 - 35 = d    ∴ d = - 2 So equation of plane is 18 x - y + 7 z = - 2 ⇒ - 18 x + y - 7 z = 2 ∴   a = - 18 ,   b = 1 , c = - 7 ,   d = 2 a + 7 b + c + 20 d = - 18 + 7 - 7 + 40 = 22

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