To solve the problem step by step, we need to find the line of intersection of the two planes given by the equations \(x + 2y + z = 6\) and \(y + 2z = 4\), and then find the foot of the perpendicular from the point \( (3, 2, 1) \) to this line.
### Step 1: Find the line of intersection of the two planes.
We have the equations:
1. \(x + 2y + z = 6\) (Equation 1)
2. \(y + 2z = 4\) (Equation 2)
From Equation 2, we can express \(y\) in terms of \(z\):
\[
y = 4 - 2z
\]
Substituting this expression for \(y\) into Equation 1:
\[
x + 2(4 - 2z) + z = 6
\]
\[
x + 8 - 4z + z = 6
\]
\[
x - 3z + 8 = 6
\]
\[
x - 3z = -2
\]
\[
x = 3z - 2
\]
Now we have \(x\) in terms of \(z\) and \(y\) in terms of \(z\):
\[
y = 4 - 2z
\]
\[
x = 3z - 2
\]
### Step 2: Parametrize the line of intersection.
Let \(z = t\). Then we can express \(x\) and \(y\) in terms of \(t\):
\[
x = 3t - 2
\]
\[
y = 4 - 2t
\]
Thus, the parametric equations of the line \(L\) are:
\[
(x, y, z) = (3t - 2, 4 - 2t, t)
\]
### Step 3: Find the foot of the perpendicular from the point \( (3, 2, 1) \) to the line.
Let \(P(\alpha, \beta, \gamma)\) be the foot of the perpendicular from the point \( (3, 2, 1) \) to the line \(L\). The coordinates of \(P\) can be expressed as:
\[
P = (3t - 2, 4 - 2t, t)
\]
The vector \(AP\) from point \(A(3, 2, 1)\) to point \(P\) is:
\[
AP = (3t - 2 - 3, 4 - 2t - 2, t - 1) = (3t - 5, 2 - 2t, t - 1)
\]
The direction vector of the line \(L\) can be derived from the parametric equations. The direction vector \(d\) is:
\[
d = \frac{d}{dt}(3t - 2, 4 - 2t, t) = (3, -2, 1)
\]
### Step 4: Set up the perpendicularity condition.
For \(AP\) to be perpendicular to \(d\), their dot product must be zero:
\[
(3t - 5, 2 - 2t, t - 1) \cdot (3, -2, 1) = 0
\]
Calculating the dot product:
\[
3(3t - 5) - 2(2 - 2t) + 1(t - 1) = 0
\]
\[
9t - 15 - 4 + 4t + t - 1 = 0
\]
\[
14t - 20 = 0
\]
\[
t = \frac{20}{14} = \frac{10}{7}
\]
### Step 5: Substitute \(t\) back to find \(P\).
Now substituting \(t = \frac{10}{7}\) into the parametric equations:
\[
\alpha = 3\left(\frac{10}{7}\right) - 2 = \frac{30}{7} - \frac{14}{7} = \frac{16}{7}
\]
\[
\beta = 4 - 2\left(\frac{10}{7}\right) = 4 - \frac{20}{7} = \frac{28}{7} - \frac{20}{7} = \frac{8}{7}
\]
\[
\gamma = \frac{10}{7}
\]
### Step 6: Calculate \(21(\alpha + \beta + \gamma)\).
Now we need to find \(\alpha + \beta + \gamma\):
\[
\alpha + \beta + \gamma = \frac{16}{7} + \frac{8}{7} + \frac{10}{7} = \frac{34}{7}
\]
Now, multiplying by 21:
\[
21(\alpha + \beta + \gamma) = 21 \cdot \frac{34}{7} = 3 \cdot 34 = 102
\]
Thus, the final answer is:
\[
\boxed{102}
\]
