3.2. $\left\{\begin{array}{l}x+y+x y=19 \\ y+z+y z=11 \\ z+x+z x=14 .\end{array}\right.$
3.2. This system can be rewritten as $x_{1} y_{1}=20, y_{1} z_{1}=$ $=12, x_{1} z_{1}=15$, where $x_{1}=x+1, y_{1}=y+1, z_{1}=z+1$. Therefore, $\left(x_{1}, y_{1}, z_{1}\right)=(5,4,3)$ or $(-5,-4,-3)$, i.e., $(x, y, z)=(4,3,2)$ or $(-6,-5,-4)$.