To solve the problem, we need to maximize the product \( xyz \) given the constraint \( 2xy + 3yz + 4xz = 18 \). We will use the method of inequalities, specifically the Arithmetic Mean-Geometric Mean (AM-GM) inequality.
### Step-by-Step Solution:
1. **State the Given Condition**:
We have the constraint:
\[
2xy + 3yz + 4xz = 18
\]
where \( x, y, z \in \mathbb{R}^+ \).
2. **Apply AM-GM Inequality**:
According to the AM-GM inequality, for non-negative real numbers \( a_1, a_2, \ldots, a_n \):
\[
\frac{a_1 + a_2 + \ldots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \ldots a_n}
\]
We can apply this to our terms \( 2xy, 3yz, 4xz \):
\[
\frac{2xy + 3yz + 4xz}{3} \geq \sqrt[3]{(2xy)(3yz)(4xz)}
\]
3. **Calculate the Right Side**:
The product on the right side is:
\[
(2xy)(3yz)(4xz) = 24x^2y^2z^2
\]
Therefore, we have:
\[
\frac{18}{3} \geq \sqrt[3]{24x^2y^2z^2}
\]
Simplifying the left side gives:
\[
6 \geq \sqrt[3]{24x^2y^2z^2}
\]
4. **Cubing Both Sides**:
Cubing both sides, we get:
\[
216 \geq 24x^2y^2z^2
\]
Dividing both sides by 24:
\[
9 \geq x^2y^2z^2
\]
Taking the square root gives:
\[
3 \geq xyz
\]
5. **Finding Maximum Value**:
The maximum value of \( xyz \) is 3, which occurs when equality holds in the AM-GM inequality. This happens when:
\[
2xy = 3yz = 4xz
\]
6. **Setting Up Equations**:
Let:
\[
2xy = k, \quad 3yz = k, \quad 4xz = k
\]
From these, we can express \( x, y, z \) in terms of \( k \):
- From \( 2xy = k \): \( y = \frac{k}{2x} \)
- From \( 3yz = k \): \( z = \frac{k}{3y} \)
- From \( 4xz = k \): \( x = \frac{k}{4z} \)
7. **Substituting**:
Substitute \( y \) and \( z \) in terms of \( x \) into the equations to solve for \( x, y, z \).
8. **Solving the Equations**:
After solving, we find:
\[
x = \frac{3}{2}, \quad y = 2, \quad z = 1
\]
9. **Final Calculation**:
Now we compute \( 2\alpha + \beta + \gamma \):
\[
2\left(\frac{3}{2}\right) + 2 + 1 = 3 + 2 + 1 = 6
\]
### Conclusion:
Thus, the value of \( 2\alpha + \beta + \gamma \) is:
\[
\boxed{6}
\]
