3. Given $\lg a<0, \lg b<0, \lg c<0$, and $\lg (a+b+c)=0$. Then the maximum value of $\lg \left(a^{2}+b^{2}+c^{2}+18 a b c\right)$ is

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3. 0 .

Given that $a, b, c$ are positive numbers less than 1, and $a+b+c=1$. We have
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$$
$\geqslant 9$ (Cauchy-Schwarz inequality).
The equality holds when $a=b=c=\frac{1}{3}$.
Transforming the inequality (1), we get
$$
\begin{array}{l}
9 a b c \leqslant a b+b c+c a \\
=\frac{1}{2}\left[(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)\right] \\
=\frac{1}{2}\left[1-\left(a^{2}+b^{2}+c^{2}\right)\right],
\end{array}
$$

which implies $a^{2}+b^{2}+c^{2}+18 a b c \leqslant 1$.
Therefore, $\lg \left(a^{2}+b^{2}+c^{2}+18 a b c\right) \leqslant 0$.
When $a=b=c=\frac{1}{3}$, $\lg \left(a^{2}+b^{2}+c^{2}+18 a b c\right)$ can achieve its maximum value of 0.