To solve the problem, we start with the given conditions about the complex numbers \( z_1, z_2, z_3 \).
### Step 1: Understanding the Complex Numbers
Since \( |z_1| = |z_2| = |z_3| = 1 \), we can express these complex numbers in polar form:
\[
z_1 = e^{i\theta_1}, \quad z_2 = e^{i\theta_2}, \quad z_3 = e^{i\theta_3}
\]
where \( \theta_1, \theta_2, \theta_3 \) are the arguments of \( z_1, z_2, z_3 \) respectively.
### Step 2: Analyzing the Given Condition
We need to analyze the condition:
\[
2(z_1 + z_2 + z_3) - 3z_1 z_2 z_3 \text{ is real.}
\]
Let \( S = z_1 + z_2 + z_3 \) and \( P = z_1 z_2 z_3 \). The expression can be rewritten as:
\[
2S - 3P
\]
This expression is real if and only if its imaginary part is zero.
### Step 3: Finding the Imaginary Part
The imaginary part of \( S \) is:
\[
\text{Im}(S) = \text{Im}(z_1) + \text{Im}(z_2) + \text{Im}(z_3) = \sin(\theta_1) + \sin(\theta_2) + \sin(\theta_3)
\]
The imaginary part of \( P \) is:
\[
\text{Im}(P) = \text{Im}(z_1 z_2 z_3) = \sin(\theta_1 + \theta_2 + \theta_3)
\]
Using the identity for the sine of a sum, we have:
\[
\sin(\theta_1 + \theta_2 + \theta_3) = \sin(\theta_1)\cos(\theta_2 + \theta_3) + \cos(\theta_1)\sin(\theta_2 + \theta_3)
\]
Thus, the imaginary part of \( 2S - 3P \) is:
\[
2(\sin(\theta_1) + \sin(\theta_2) + \sin(\theta_3)) - 3\sin(\theta_1 + \theta_2 + \theta_3)
\]
Setting this equal to zero gives us the condition we need to satisfy.
### Step 4: Finding Maximum Argument
We need to find the maximum of \( \arg(z_1), \arg(z_2), \arg(z_3) \) under the constraint that they are positive. Let:
\[
M = \max(\arg(z_1), \arg(z_2), \arg(z_3))
\]
We want to find the minimum value of \( M \) such that the condition holds.
### Step 5: Evaluating Cases
We can evaluate specific cases for the angles \( \theta_1, \theta_2, \theta_3 \). The angles can be expressed in terms of \( \frac{k\pi}{6} \) where \( k \) is an integer.
1. If \( \theta_1 = \frac{\pi}{3} \), \( \theta_2 = \frac{\pi}{3} \), \( \theta_3 = \frac{\pi}{3} \), then \( M = \frac{\pi}{3} \).
2. If \( \theta_1 = \frac{\pi}{6} \), \( \theta_2 = \frac{\pi}{6} \), \( \theta_3 = \frac{\pi}{6} \), then \( M = \frac{\pi}{6} \).
### Step 6: Conclusion
The minimum value of \( M \) occurs when \( M = \frac{\pi}{6} \). Therefore, we have:
\[
k = 1 \quad \text{(since } M = \frac{k\pi}{6} \text{)}
\]
Thus, \( k + 2 = 1 + 2 = 3 \).
### Final Answer
The value of \( k + 2 \) is \( 3 \).
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