7. Real numbers $x, y, z$ satisfy
$$
x+y+z=1 \text {, and } x^{2}+y^{2}+z^{2}=3 \text {. }
$$

Then the range of $x y z$ is $\qquad$
7. $\left[-1, \frac{5}{27}\right]$.

Let $f=x y z$. First, transform and reduce the variables to a single-variable function.
Notice,
$$
\begin{aligned}
1 & =(x+y+z)^{2} \\
& =x^{2}+y^{2}+z^{2}+2(x y+y z+z x) \\
& =3+2(x y+y z+z x) \\
& \Rightarrow x y+y z+z x=-1 \\
& \Rightarrow x^{2}(y+z)+x y z=-x \\
& \Rightarrow f(x)=-x-x^{2}(1-x)=x^{3}-x^{2}-x \\
& \Rightarrow f^{\prime}(x)=3 x^{2}-2 x-1 .
\end{aligned}
$$

By the Cauchy-Schwarz inequality,
$$
\begin{array}{l}
2\left(y^{2}+z^{2}\right) \geqslant(y+z)^{2} \\
\Rightarrow 2\left(3-x^{2}\right) \geqslant(1-x)^{2} \\
\Rightarrow-1 \leqslant x \leqslant \frac{5}{3} .
\end{array}
$$

Let $f^{\prime}(x)=0$, we get $x=1,-\frac{1}{3}$.
Thus, when $(x, y, z)=\left(-\frac{1}{3},-\frac{1}{3}, \frac{5}{3}\right)$,
$$
\begin{array}{l}
f_{\text {max }}=\frac{5}{27} ; \\
\text { When }(x, y, z)=(1,1,-1) \text {, } \\
f_{\min }=-1 .
\end{array}
$$