10. Let $a, b, c$ all be greater than 1, satisfying $\left\{\begin{array}{l}\lg a+\log _{b} c=3, \\ \lg b+\log _{a} c=4 .\end{array}\right.$ Find the maximum value of $\lg a \cdot \lg c$.
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ \operatorname { l g } a + \frac { \operatorname { l g } c } { \operatorname { l g } b } = 3 , } \\
{ \operatorname { l g } b + \frac { \operatorname { l g } c } { \operatorname { l g } a } = 4 }
\end{array} \Rightarrow \left\{\begin{array}{l}
\lg a \cdot \lg b+\lg c=3 \lg b, \\
\lg a \cdot \lg b+\lg c=4 \lg a
\end{array} \Rightarrow \lg b=\frac{4}{3} \lg a \Rightarrow \frac{4}{3} \lg a+\frac{\lg c}{\lg a}=4\right.\right. \\
\Rightarrow 4 \lg ^{2} a+3 \lg c=12 \lg a \Rightarrow \lg c=\frac{4}{3}\left(3 \lg a-\lg ^{2} a\right) \Rightarrow \lg a \cdot \lg c=\frac{4}{3}\left(3 \lg ^{2} a-\lg ^{3} a\right) .
\end{array}
$$

According to the problem, $\lg a, \lg c>0 \Rightarrow 0<\lg a<3$,
Let $f(x)=3 x^{2}-x^{3}, x \in(0,3)$, then $f^{\prime}(x)=6 x-3 x^{2}=3 x(2-x)$,
Thus $f(x)$ is monotonically increasing on $(0,2)$ and monotonically decreasing on $(2,3)$,
Therefore, $f(x)_{\max }=f(2)=3 \times 2^{2}-2^{3}=4$.
In summary, when $\lg a=2, \lg b=\lg c=\frac{8}{3}$, the maximum value of $\lg a \cdot \lg c$ is $\frac{16}{3}$.
$$