Foot of Perpendicular from Point to Plane

To find the foot of the perpendicular from the point $P(1, -2, 1)$ to the given plane, we first need to determine the value of the constant $\alpha$.

The equation of the plane is:
$x + 2y - 2z = \alpha \quad \implies \quad x + 2y - 2z - \alpha = 0$

The perpendicular distance $d$ from a point $(x_1, y_1, z_1)$ to a plane $Ax + By + Cz + D = 0$ is given by:
$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Given $P(x_1, y_1, z_1) = (1, -2, 1)$, $d = 5$, and $\alpha > 0$:
$5 = \frac{|1(1) + 2(-2) - 2(1) - \alpha|}{\sqrt{1^2 + 2^2 + (-2)^2}}$

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

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

$15 = |5 + \alpha|$

Since $\alpha > 0$, we have $5 + \alpha > 0$, which gives:
$5 + \alpha = 15 \implies \alpha = 10$

So, the equation of the plane is:
$x + 2y - 2z - 10 = 0$

Now, let $Q(x, y, z)$ be the foot of the perpendicular from $P(1, -2, 1)$ to the plane $x + 2y - 2z - 10 = 0$.

Using the formula for the foot of the perpendicular:
$\frac{x - x_1}{A} = \frac{y - y_1}{B} = \frac{z - z_1}{C} = -\frac{Ax_1 + By_1 + Cz_1 + D}{A^2 + B^2 + C^2}$

Substituting the values:
$\frac{x - 1}{1} = \frac{y - (-2)}{2} = \frac{z - 1}{-2} = -\frac{1(1) + 2(-2) - 2(1) - 10}{1^2 + 2^2 + (-2)^2}$

$\frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 1}{-2} = -\frac{-15}{9} = \frac{5}{3}$

Solving for $x$, $y$, and $z$:

1. For $x$:
   $\frac{x - 1}{1} = \frac{5}{3} \implies x = 1 + \frac{5}{3} = \frac{8}{3}$

2. For $y$:
   $\frac{y + 2}{2} = \frac{5}{3} \implies y + 2 = \frac{10}{3} \implies y = \frac{10}{3} - 2 = \frac{4}{3}$

3. For $z$:
   $\frac{z - 1}{-2} = \frac{5}{3} \implies z - 1 = -\frac{10}{3} \implies z = 1 - \frac{10}{3} = -\frac{7}{3}$

Thus, the foot of the perpendicular is $\left(\frac{8}{3}, \frac{4}{3}, -\frac{7}{3}\right)$.

Correct Option: A