Given that $\tan \alpha = 3$, we need to find the value of the expression $\frac{\sin \alpha - \cos \alpha}{2\sin \alpha + 3\cos \alpha}$.

1. Problem Description

Given that $\tan \alpha = 3$, we need to find the value of the expression $\frac{\sin \alpha - \cos \alpha}{2\sin \alpha + 3\cos \alpha}$.

2. Solution Steps

We can rewrite the expression in terms of $\tan \alpha$ by dividing both the numerator and the denominator by $\cos \alpha$:

$\frac{\sin \alpha - \cos \alpha}{2\sin \alpha + 3\cos \alpha} = \frac{\frac{\sin \alpha}{\cos \alpha} - \frac{\cos \alpha}{\cos \alpha}}{2\frac{\sin \alpha}{\cos \alpha} + 3\frac{\cos \alpha}{\cos \alpha}} = \frac{\tan \alpha - 1}{2\tan \alpha + 3}$

Now, we substitute $\tan \alpha = 3$ into the expression:

$\frac{3 - 1}{2(3) + 3} = \frac{2}{6 + 3} = \frac{2}{9}$

3. Final Answer

The value of the expression is $\frac{2}{9}$. So the answer is [C].