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

$\left(\frac{8}{3}, \frac{4}{3}, \frac{-7}{3}\right)$

We have, 

PM = 5

$⇒ \frac{|1-4-2-\alpha |}{\sqrt{1+4+4}}=5$

$⇒ |\alpha + 5| = 15 $

$⇒ \alpha + 5 = ± 15$

$⇒  \alpha = 10, -20 $

$⇒\alpha = 10$             $[∵ \alpha > 0 ]$

The equation of PM is

$\frac{x-1}{1}=\frac{y+2}{2}=\frac{z-1}{-2}$

The coordinates of M are given by

$\frac{x-1}{1}=\frac{y+2}{2}=\frac{z-1}{-2}=r$

So, the coordinates of M are (r + 1, 2r - 2, -2r + 1). As M lies on the plane $x + 2y - 2z = \alpha .$

$∴ r + 1 + 4r - 4 + 4r - 2 = 10 $              $[∵ \alpha = 10 ]$

$⇒ 9r = 15 ⇒ r = \frac{5}{3}$

Hence, the coordinates of M are $\left(\frac{8}{3}, \frac{4}{3}, \frac{-7}{3}\right)$