6. The positive integer solution of the system of equations $\left\{\begin{array}{l}\frac{z^{y}}{y}=y^{2 x}, \\ 2^{z}=2 \cdot 4^{x}, \\ x+y+z=12\end{array}\right.$ is $\qquad$
$6 . x=2, y=5, z=5$.
From $2^{z}=2^{2 x+1} \Rightarrow z=2 x+1$. Substituting into $x+y+z=12 \Rightarrow 3 x+y=11$.
$\because x, y$ are positive integers
then $\left\{\begin{array}{l}x=1, \\ y=8\end{array}\right.$ or $\left\{\begin{array}{l}x=2, \\ y=5\end{array}\right.$ or $\left\{\begin{array}{l}x=3, \\ y=2 .\end{array}\right.$ Substituting into $z^{y}=y^{2 x+1}$, we get $z=5$. That is, $x=2, y=5, z=5$ is the solution.