7. Given real numbers $a, b, c$ satisfy $a+b+c=1, a^{2}+b^{2}+c^{2}=3$. Then the maximum value of $a b c$ is $\qquad$ .
7. $\frac{5}{27}$.

Notice that among $a^{2}, b^{2}, c^{2}$, there is always one that does not exceed 1, let's assume $c^{2} \leqslant 1$. Then $c \in[-1,1]$.
$$
\begin{array}{l}
\text { Also } a b c=\frac{1}{2}\left[(a+b)^{2}-\left(a^{2}+b^{2}\right)\right] c \\
=\frac{1}{2}\left[(1-c)^{2}-\left(3-c^{2}\right)\right] c=c^{3}-c^{2}-c . \\
\text { Let } f(x)=x^{3}-x^{2}-x(x \in[-1,1]) .
\end{array}
$$

Then by $f^{\prime}(x)=3 x^{2}-2 x-1=0$, we get
$$
x_{1}=-\frac{1}{3}, x_{2}=1 \text {. }
$$

It is easy to see that $f(x)_{\max }=f\left(-\frac{1}{3}\right)=\frac{5}{27}$.