Solve
$$
x+2 x y+3 x y z=x(1+2 y+3 y z)=115=5 \times 23 \text{, }
$$

When $x=1$, $2 y+3 y z=y(2+3 z)=114=2 \times 3 \times 19$.
Clearly, $2+3 z \geqslant 5$, and it is congruent to 2 modulo 3, then $\left\{\begin{array}{l}2+3 z=38, \\ y=3\end{array} \Rightarrow\left\{\begin{array}{l}y=3, \\ z=12 ;\end{array}\right.\right.$
When $x=5$, then $2 y+3 y z=2 \times 11 \Rightarrow\left\{\begin{array}{l}2+3 z=11, \\ y=2\end{array} \Rightarrow\left\{\begin{array}{l}y=2, \\ z=3 .\end{array}\right.\right.$
In summary, $x+y+z=1+3+12=16$ or $x+y+z=5+2+3=10$.
6. If positive integers $x, y, z$ satisfy $x+2 x y+3 x y z=115$, then $x+y+z=$