To solve the problem, we need to follow these steps:
### Step 1: Find the point where the line intersects the plane
The line is given by the symmetric equations:
\[
\frac{x-2}{2} = \frac{y+2}{2} = \frac{z-1}{1}
\]
We can express the coordinates of any point on this line in terms of a parameter \( r \):
\[
x = 2r + 2, \quad y = 2r - 2, \quad z = r + 1
\]
Now, we need to find the value of \( r \) when this point lies on the plane defined by:
\[
x + y + z = 21
\]
Substituting the expressions for \( x, y, z \):
\[
(2r + 2) + (2r - 2) + (r + 1) = 21
\]
Simplifying this:
\[
2r + 2r + r + 2 - 2 + 1 = 21
\]
\[
5r + 1 = 21
\]
Subtracting 1 from both sides:
\[
5r = 20
\]
Dividing by 5:
\[
r = 4
\]
### Step 2: Calculate the coordinates of the intersection point
Now substituting \( r = 4 \) back into the equations for \( x, y, z \):
\[
x = 2(4) + 2 = 8 + 2 = 10
\]
\[
y = 2(4) - 2 = 8 - 2 = 6
\]
\[
z = 4 + 1 = 5
\]
Thus, the point of intersection is \( (10, 6, 5) \).
### Step 3: Find the equation of the plane P
The plane \( P \) is perpendicular to the line. The direction ratios of the line are \( (2, 2, 1) \). Therefore, the normal vector to the plane \( P \) can be taken as \( (2, 2, 1) \).
The general equation of the plane can be written as:
\[
2x + 2y + z + k = 0
\]
To find \( k \), we substitute the point \( (10, 6, 5) \) into the plane equation:
\[
2(10) + 2(6) + 5 + k = 0
\]
Calculating:
\[
20 + 12 + 5 + k = 0
\]
\[
37 + k = 0 \implies k = -37
\]
Thus, the equation of the plane \( P \) is:
\[
2x + 2y + z - 37 = 0
\]
### Step 4: Calculate the distance from the origin to the plane
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 the origin \( (0, 0, 0) \) and the plane \( 2x + 2y + z - 37 = 0 \):
\[
D = \frac{|2(0) + 2(0) + 1(0) - 37|}{\sqrt{2^2 + 2^2 + 1^2}} = \frac{|-37|}{\sqrt{4 + 4 + 1}} = \frac{37}{\sqrt{9}} = \frac{37}{3}
\]
### Final Answer
The distance of plane \( P \) from the origin is:
\[
\frac{37}{3} \text{ units}
\]
