If the distance of the point ` P (1, -2,1)` from the plane `x +2 y -2 z =\alpha`, where `\alpha > 0` is 5, then the foot

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Q 1619180910.     If the distance of the point ` P (1, -2,1)` from the plane `x +2 y -2 z =\alpha`, where `\alpha > 0` is 5, then the foot of the perpendicular from `P` is
JEE 2010 ADVANCED Paper 2
A

`( \frac{8}{3}, \frac{4}{3}, -\frac{7}{3})`

B

`( \frac{4}{3}, -\frac{4}{3}, \frac{1}{3})`

C

` ( \frac{1}{3}, \frac{2}{3}, \frac{10}{3})`

D

`( \frac{2}{3}, -\frac{1}{3}, \frac{5}{2})`

HINT


Solution
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

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