6. Given that $x, y, z$ are positive real numbers, and $x+y+z=1$. If $\frac{a}{x y z}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-2$, then the range of the real number $a$ is . $\qquad$
6. $\left(0, \frac{7}{27}\right]$.

It is easy to see that $x, y, z \in (0,1)$, and $1-x, 1-y, 1-z \in (0,1)$.
On one hand, it is easy to get $a = xy + yz + zx - 2xyz > xy > 0$.
When $x, y \rightarrow 0, z \rightarrow 1$, $a \rightarrow 0$. On the other hand, it is easy to get $a = (1-x)x + (1-2x)yz$.
(1) When $0 < x < \frac{1}{2}$, since $y, z \in (0,1)$, we have $(1-2x)yz > 0$, so $a < (1-x)x \leq \frac{1}{4}$.
In summary, we have $a \in \left(0, \frac{7}{27}\right]$.