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

Distance of point $P(l, -2, 1)$ from plane $x+2y-2z-\alpha =0\text{ is }5\text{. }$

$\Rightarrow \frac{|1-4-2-\alpha |}{\sqrt{1+4+4}}=5\Rightarrow |\alpha +5|=15\Rightarrow \alpha =10$

Let $Q$ be the foot of perpendicluar from $P$ on the plane 

equation of the line $PQ$ is  $\frac{x-1}{1}=\frac{y+2}{2}=\frac{z-1}{-2}=r$

Coordinates of point $Q$ are $(r+1,2r-2,1-2r)$

Since it lies on the plane

$\begin{array}{l}(r+1)+2(2r-2)-2(1-2r)=10\\ \Rightarrow 9r=15\text{ or }r=5/3\\ \therefore Q\equiv \left(\frac{8}{3},\frac{4}{3},\frac{-7}{3}\right)\end{array}$