To find the distance of the point \( P(-2, 1, 2) \) from the line of intersection of the planes given by the equations \( x + 3y - 2z + 1 = 0 \) and \( x - 2y + z = 0 \), we can follow these steps:
### Step 1: Find the Direction Ratios of the Line of Intersection
The direction ratios of the line of intersection can be found using the cross product of the normals of the two planes.
1. The normal vector of the first plane \( x + 3y - 2z + 1 = 0 \) is \( \mathbf{n_1} = (1, 3, -2) \).
2. The normal vector of the second plane \( x - 2y + z = 0 \) is \( \mathbf{n_2} = (1, -2, 1) \).
Now, we calculate the cross product \( \mathbf{n_1} \times \mathbf{n_2} \):
\[
\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 3 & -2 \\
1 & -2 & 1
\end{vmatrix}
\]
Calculating the determinant:
\[
\mathbf{d} = \mathbf{i} \left( 3 \cdot 1 - (-2) \cdot (-2) \right) - \mathbf{j} \left( 1 \cdot 1 - (-2) \cdot 1 \right) + \mathbf{k} \left( 1 \cdot (-2) - 3 \cdot 1 \right)
\]
\[
= \mathbf{i} (3 - 4) - \mathbf{j} (1 + 2) + \mathbf{k} (-2 - 3)
\]
\[
= -\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}
\]
Thus, the direction ratios of the line of intersection are \( (-1, -3, -5) \).
### Step 2: Find a Point on the Line of Intersection
To find a point on the line of intersection, we can set \( z = 0 \) and solve the equations:
1. From \( x + 3y + 1 = 0 \) (setting \( z = 0 \)):
\[
x + 3y = -1 \quad (1)
\]
2. From \( x - 2y = 0 \):
\[
x = 2y \quad (2)
\]
Substituting (2) into (1):
\[
2y + 3y = -1 \implies 5y = -1 \implies y = -\frac{1}{5}
\]
Now substituting \( y \) back to find \( x \):
\[
x = 2 \left(-\frac{1}{5}\right) = -\frac{2}{5}
\]
Thus, a point on the line is \( Q\left(-\frac{2}{5}, -\frac{1}{5}, 0\right) \).
### Step 3: Find the Distance from Point \( P \) to Line
The distance \( d \) from point \( P \) to the line can be calculated using the formula:
\[
d = \frac{|\mathbf{PQ} \cdot \mathbf{d}|}{|\mathbf{d}|}
\]
Where \( \mathbf{PQ} \) is the vector from point \( Q \) to point \( P \):
\[
\mathbf{PQ} = P - Q = \left(-2 + \frac{2}{5}, 1 + \frac{1}{5}, 2 - 0\right) = \left(-\frac{8}{5}, \frac{6}{5}, 2\right)
\]
Now, we compute the dot product \( \mathbf{PQ} \cdot \mathbf{d} \):
\[
\mathbf{d} = (-1, -3, -5)
\]
Calculating the dot product:
\[
\mathbf{PQ} \cdot \mathbf{d} = \left(-\frac{8}{5}\right)(-1) + \left(\frac{6}{5}\right)(-3) + (2)(-5)
\]
\[
= \frac{8}{5} - \frac{18}{5} - 10 = \frac{8 - 18 - 50}{5} = \frac{-60}{5} = -12
\]
### Step 4: Calculate the Magnitude of \( \mathbf{d} \)
The magnitude of \( \mathbf{d} \):
\[
|\mathbf{d}| = \sqrt{(-1)^2 + (-3)^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}
\]
### Step 5: Calculate the Distance
Now substituting back into the distance formula:
\[
d = \frac{|-12|}{\sqrt{35}} = \frac{12}{\sqrt{35}}
\]
Thus, the distance of the point \( P(-2, 1, 2) \) from the line of intersection of the given planes is:
\[
\boxed{\frac{12}{\sqrt{35}}}
\]
