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
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

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

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