7. If real numbers $a, b, c$ satisfy
$$
a+2 b+3 c=6, a^{2}+4 b^{2}+9 c^{2}=12 \text {, }
$$

then $a b c=$ . $\qquad$
7. $\frac{4}{3}$.

Solution 1 From the given, we have
$$
\begin{array}{l}
\left(a^{2}+4 b^{2}+9 c^{2}\right)-4(a+2 b+3 c)=12-4 \times 6 \\
\Rightarrow(a-2)^{2}+4(b-1)^{2}+(3 c-2)^{2}=0 \\
\Rightarrow a=2, b=1, c=\frac{2}{3} \\
\Rightarrow a b c=\frac{4}{3} .
\end{array}
$$

Solution 2 From the given, we have
$$
\begin{array}{l}
a+2 b=6-3 c, a^{2}+4 b^{2}=12-9 c^{2} . \\
\text { By }\left(\frac{a+2 b}{2}\right)^{2} \leqslant \frac{a^{2}+4 b^{2}}{2} \\
\Rightarrow\left(\frac{6-3 c}{2}\right)^{2} \leqslant \frac{12-9 c^{2}}{2} \\
\Rightarrow(3 c-2)^{2} \leqslant 0 \\
\Rightarrow 3 c-2=0 \Rightarrow c=\frac{2}{3} .
\end{array}
$$

Thus, $a+2 b=4, a^{2}+4 b^{2}=8$.
$$
\begin{array}{l}
\text { Also, } 8=a^{2}+4 b^{2}=(a+2 b)^{2}-4 a b=16-4 a b \\
\Rightarrow a b=2 .
\end{array}
$$

Therefore, $a b c=\frac{4}{3}$.
Solution 3 From the given and Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
36=(a+2 b+3 c)^{2} \\
\leqslant\left(1^{2}+1^{2}+1^{2}\right)\left(a^{2}+4 b^{2}+9 c^{2}\right)=36 .
\end{array}
$$

Equality holds if and only if $a=2 b=3 c=\frac{6}{3}$, i.e., $a=2, b=1, c=\frac{2}{3}$.
Therefore, $a b c=\frac{4}{3}$.