To solve the problem, we need to find the foot of the perpendicular from the point \( P(1, -2, 1) \) to the plane given by the equation \( x + 2y - 2z = \alpha \), where \( \alpha > 0 \) and the distance from point \( P \) to the plane is 5.
### Step 1: Understanding the Distance from a Point to a Plane
The formula for the distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[
d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]
For our plane \( x + 2y - 2z = \alpha \), we can rewrite it as:
\[
x + 2y - 2z - \alpha = 0
\]
Here, \( A = 1 \), \( B = 2 \), \( C = -2 \), and \( D = -\alpha \).
### Step 2: Plugging in the Point Coordinates
Now we plug in the coordinates of point \( P(1, -2, 1) \):
\[
d = \frac{|1 \cdot 1 + 2 \cdot (-2) - 2 \cdot 1 - \alpha|}{\sqrt{1^2 + 2^2 + (-2)^2}}
\]
Calculating the denominator:
\[
\sqrt{1 + 4 + 4} = \sqrt{9} = 3
\]
Now substituting into the distance formula:
\[
d = \frac{|1 - 4 - 2 - \alpha|}{3} = \frac{|-5 - \alpha|}{3}
\]
### Step 3: Setting the Distance Equal to 5
According to the problem, this distance is given to be 5:
\[
\frac{|-5 - \alpha|}{3} = 5
\]
Multiplying both sides by 3:
\[
|-5 - \alpha| = 15
\]
This gives us two cases:
1. \( -5 - \alpha = 15 \)
2. \( -5 - \alpha = -15 \)
### Step 4: Solving the Cases
**Case 1:**
\[
-5 - \alpha = 15 \implies -\alpha = 20 \implies \alpha = -20 \quad (\text{not valid since } \alpha > 0)
\]
**Case 2:**
\[
-5 - \alpha = -15 \implies -\alpha = -10 \implies \alpha = 10 \quad (\text{valid})
\]
### Step 5: Finding the Foot of the Perpendicular
Now that we have \( \alpha = 10 \), the equation of the plane becomes:
\[
x + 2y - 2z = 10
\]
The foot of the perpendicular from point \( P(1, -2, 1) \) to the plane can be found using the direction ratios given by the normal vector of the plane, which is \( (1, 2, -2) \).
### Step 6: Equation of the Line
The parametric equations of the line passing through \( P \) and perpendicular to the plane are:
\[
\frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 1}{-2} = t
\]
From these, we can express \( x, y, z \) in terms of \( t \):
\[
x = 1 + t, \quad y = -2 + 2t, \quad z = 1 - 2t
\]
### Step 7: Substitute into the Plane Equation
Substituting these into the plane equation \( x + 2y - 2z = 10 \):
\[
(1 + t) + 2(-2 + 2t) - 2(1 - 2t) = 10
\]
Simplifying:
\[
1 + t - 4 + 4t - 2 + 4t = 10
\]
\[
9t - 5 = 10 \implies 9t = 15 \implies t = \frac{5}{3}
\]
### Step 8: Finding the Coordinates of the Foot
Now substituting \( t = \frac{5}{3} \) back into the parametric equations:
\[
x = 1 + \frac{5}{3} = \frac{8}{3}, \quad y = -2 + 2 \cdot \frac{5}{3} = \frac{4}{3}, \quad z = 1 - 2 \cdot \frac{5}{3} = -\frac{7}{3}
\]
Thus, the coordinates of the foot of the perpendicular are:
\[
\left( \frac{8}{3}, \frac{4}{3}, -\frac{7}{3} \right)
\]
### Conclusion
The foot of the perpendicular from the point \( P(1, -2, 1) \) to the plane \( x + 2y - 2z = 10 \) is:
\[
\boxed{\left( \frac{8}{3}, \frac{4}{3}, -\frac{7}{3} \right)}
\]
