3.1. $\left\{\begin{array}{l}x(y+z)=35 \\ y(x+z)=32 \\ z(x+y)=27 .\end{array}\right.$
3.1. The system is linear with respect to the unknowns $x_{1}=y z$, $y_{1}=x z, z_{1}=x y$. To find $x_{1}$, we need to add the last two equations and subtract the first equation from them. The result is $x_{1}=(32+27-35) / 2=12$. After this, we find $y_{1}=15$ and $z_{1}=20$.

Thus, $y z=12, x z=15, x y=20$. Therefore, $x^{2}=\frac{(x y)(x z)}{y z}=\frac{20 \cdot 15}{12}=$ $=25$, i.e., $x= \pm 5$. Moreover, $y=20 / x= \pm 4$ and $z=15 / x= \pm 3$. In the end, we get two solutions ( $5,4,3$ ) and ( $-5,-4,-3$ ).