The length of the foot of the perpendicular from the point (1, 3 2 , 2 ) to the plane 2 x-2 y+4 z+17=0 is

Step-by-step solution

The perpendicular distance from point P(x₁, y₁, z₁) to the plane Ax + By + Cz + D = 0 is given by d = |Ax₁ + By₁ + Cz₁ + D| A^2 + B^2 + C^2 . For point P (1, 3 2 , 2 ) and plane 2x - 2y + 4z + 17 = 0 , we substitute A = 2 , B = -2 , C = 4 , D = 17 . Evaluating the numerator: |2(1) - 2 ( 3 2 ) + 4(2) + 17| = |2 - 3 + 8 + 17| = |24| = 24 . The denominator simplifies to: 2^2 + (-2)^2 + 4^2 = 4 + 4 + 16 = 24 = 2 6 . Thus, the distance is d = 24 2 6 = 12 6 = 12 6 6 = 2 6 . This result corresponds to option D .