NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice

A plane passes through 1 , - 2 , 1 and is perpendicular to two planes 2 x - 2 y + z = 0 and x - y + 2 z = 4 . The distance of the plane from the point 0 , 2 , 2 is

Options

A3 2 units
B4 2 units
C3 2 units
D2 2 units

Correct answer

A. 3 2 units

Step-by-step solution

Normal vector perpendicular to the required plane is i ^ j ^ k ^ 2 - 2 1 1 - 1 2 = i ^ - 3 - j ^ 3 + k ^ 0 = - 3 i ^ - 3 j ^ Equation of the plane through 1 , - 2 , 1 is 1 x - 1 + 1 y + 2 + 0 z - 1 = 0 x + y + 1 = 0 Distance from 0 , 2 , 2 is 2 + 1 2 = 3 2

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