7. Given that $x, y, z$ are positive real numbers, satisfying $x+y+z=1$, 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
7. $\left(0, \frac{7}{27}\right]$

Analysis: From $\frac{a}{x y z}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-2$,
we know that $a=x y+y z+z x-2 x y z=(x y+y z+z x)(x+y+z)-2 x y z=x y z+\sum_{\text {sym }} x^{2} y$
On one hand, by Schur's inequality, we have: $\sum x^{3}+3 x y z \geq \sum\left(x^{2} y+x y^{2}\right)$,
thus $x y z+\sum_{\text {sym }} x^{2} y \leq x y z+\frac{(x+y+z)^{3}-3 x y z}{4}=\frac{1+x y z}{4} \leq \frac{1+\frac{1}{27}}{4}=\frac{7}{27}$, with equality holding if and only if $x=y=z$.

On the other hand, since $x, y, z$ are all positive real numbers, $x y z+\sum_{\text {sym }} x^{2} y>0$, and as $x \rightarrow 0, y \rightarrow 0, z \rightarrow 1$, it approaches 0. Therefore, the range of $a$ is $\left(0, \frac{7}{27}\right]$.