5. (Mathematics of Intermediate Level, Issue 1, 2005 Olympiad Training Problem) 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 $\qquad$ .
5. Fill $\left(0, \frac{7}{27}\right]$. Reason: It is easy to know 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 $a = (1-x)x + (1-2x)yz$.
When $0 < x < 1$, $(1-2x)yz > 0$, so, $a < (1-x)x \leqslant \frac{1}{4}$.
In summary, we have $a \in \left(0, \frac{7}{27}\right]$.