Solution:
Let $x + 2y + 3z = \frac{11}{12}$.

Using the notations $a = bx$, $b = 3$, and $c = 2$, from the equation we obtain the equality
$$2a + 8b + 9c = 11.$$

We also have
$$a b c + a b + a c + b c + a + b + c + 1 = (a + 1)(b + 1)(c + 1) \times \frac{125}{18}.$$

By using the inequality
$$\begin{gathered} 14(a + 1 + b - 1) = (2c + 2) + 6 + 8) 0c + 9 = \\ \left.\frac{2a + 2 + b + 8 + 9c + 9}{3}\right) = \left(\frac{2a + 8b + 9c + 1}{3}\right) = 10 \end{gathered}$$

From the basic inequality, it follows that $(a + 1)(b + 1)(c + 1) \leq \frac{125}{18}$.

Equality holds if $2a + 8b + 9c = 10$ or $a = 4, b = \frac{1}{4}$, $c = \frac{1}{9}$, that is $x = \frac{2}{3}$, $y = \frac{1}{12}$, $z = \frac{1}{30}$.