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

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Q 1347112983.     If the distance of the point `P(1,-2,1)` from the plane `x+2y-2z=a`, where `a>0`, is `5`, then the foot of the `perp` from `P` to this plane is ?
A

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

B

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

C

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

D

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

HINT


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

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