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AIEEE 2010

JEE Main 2010 Mathematics question on Three Dimensional Geometry

If the distance of the point P(1,−2,1)P(1, -2, 1) from the plane x+2y−2z=αx + 2y - 2z = \alpha, where α>0\alpha > 0, is 5, then the foot of the perpendicular from PP to the plane is
  1. A.
    (83,43,−73)\left(\dfrac{8}{3}, \dfrac{4}{3}, -\dfrac{7}{3}\right)
  2. B.
    (43,−43,13)\left(\dfrac{4}{3}, -\dfrac{4}{3}, \dfrac{1}{3}\right)
  3. C.
    (13,23,103)\left(\dfrac{1}{3}, \dfrac{2}{3}, \dfrac{10}{3}\right)
  4. D.
    (23,−13,52)\left(\dfrac{2}{3}, -\dfrac{1}{3}, \dfrac{5}{2}\right)
Show answer

Correct option: A

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