To solve the problem, we need to find the value of \( \alpha + 2\beta \) given that the point \( (\alpha, \beta, 2) \) lies on a line through the point \( (3, 6, -2) \) and is parallel to the planes defined by the equations \( x - y + 2z = 5 \) and \( 3x + y + 2z = 6 \).
### Step-by-Step Solution:
1. **Find the Normal Vectors of the Planes**:
- For the plane \( x - y + 2z = 5 \), the normal vector \( \mathbf{n_1} \) is given by the coefficients of \( x, y, z \):
\[
\mathbf{n_1} = (1, -1, 2)
\]
- For the plane \( 3x + y + 2z = 6 \), the normal vector \( \mathbf{n_2} \) is:
\[
\mathbf{n_2} = (3, 1, 2)
\]
2. **Find a Direction Vector Parallel to the Line**:
- The direction vector \( \mathbf{d} \) of the line can be found using the cross product of the two normal vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \):
\[
\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2}
\]
- Calculating the cross product:
\[
\mathbf{d} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & -1 & 2 \\
3 & 1 & 2
\end{vmatrix}
\]
- This determinant expands to:
\[
\mathbf{d} = \mathbf{i}((-1)(2) - (2)(1)) - \mathbf{j}((1)(2) - (2)(3)) + \mathbf{k}((1)(1) - (-1)(3))
\]
\[
= \mathbf{i}(-2 - 2) - \mathbf{j}(2 - 6) + \mathbf{k}(1 + 3)
\]
\[
= -4\mathbf{i} + 4\mathbf{j} + 4\mathbf{k}
\]
- Thus, the direction vector is:
\[
\mathbf{d} = (-4, 4, 4)
\]
3. **Equation of the Line**:
- The line passing through the point \( (3, 6, -2) \) with direction vector \( (-4, 4, 4) \) can be parametrized as:
\[
\begin{align*}
x &= 3 - 4t \\
y &= 6 + 4t \\
z &= -2 + 4t
\end{align*}
\]
4. **Substituting the Point**:
- We need to find \( \alpha \) and \( \beta \) such that the point \( (\alpha, \beta, 2) \) lies on this line. Set \( z = 2 \):
\[
-2 + 4t = 2 \implies 4t = 4 \implies t = 1
\]
- Substitute \( t = 1 \) into the equations for \( x \) and \( y \):
\[
x = 3 - 4(1) = 3 - 4 = -1 \implies \alpha = -1
\]
\[
y = 6 + 4(1) = 6 + 4 = 10 \implies \beta = 10
\]
5. **Calculate \( \alpha + 2\beta \)**:
- Now, we can find \( \alpha + 2\beta \):
\[
\alpha + 2\beta = -1 + 2(10) = -1 + 20 = 19
\]
### Final Answer:
Thus, the value of \( \alpha + 2\beta \) is \( \boxed{19} \).
