1. Given data

Point: $P(1,-2,1)$

Plane: $x+2y-2z=\alpha, \qquad \alpha>0$

Distance of point from plane is $5$.

2. Use distance formula

For plane $ax+by+cz+d=0$ and point $(x_1,y_1,z_1)$, distance is $\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}.$

Here, $x+2y-2z-\alpha=0$ so $a=1,\; b=2,\; c=-2,\; d=-\alpha.$

Substitute $P(1,-2,1)$: $\frac{|1+2(-2)-2(1)-\alpha|}{\sqrt{1^2+2^2+(-2)^2}}=5$

$\frac{|1-4-2-\alpha|}{\sqrt{9}}=5$

$\frac{| -5-\alpha |}{3}=5$

$|-5-\alpha|=15$

So, $-5-\alpha=\pm 15$

Case 1: $-5-\alpha=15 \implies \alpha=-20$ Rejected since $\alpha>0$.

Case 2: $-5-\alpha=-15 \implies \alpha=10$

Hence the plane is $x+2y-2z=10.$

3. Find foot of perpendicular from $P$ to the plane

The normal vector of the plane is $\vec n=(1,2,-2).$

If the foot of perpendicular is $F$, then $F=P+t\vec n=(1+t,-2+2t,1-2t).$

Since $F$ lies on the plane, $x+2y-2z=10.$

Substitute: $(1+t)+2(-2+2t)-2(1-2t)=10$

$1+t-4+4t-2+4t=10$

$9t-5=10$

$9t=15$

$t=\frac{5}{3}.$

Therefore, $F=\left(1+\frac53,\,-2+2\cdot\frac53,\,1-2\cdot\frac53\right)$

$F=\left(\frac83,\frac43,-\frac73\right).$

4. Match with options

This is $\left(\frac83,\frac43,-\frac73\right),$ which is Option A.

5. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.