11. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1, a_{2}=2, a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+a_{n+2}$, and $a_{n+1} a_{n+2} \neq 1, S_{1999}=\sum_{n=1}^{1999} a_{n}=$ $\qquad$
11. 3997 .
$a_{3}=\frac{a_{1}+a_{2}}{a_{1} a_{2}-1}=3, a_{4}=\frac{a_{2}+a_{3}}{a_{2} a_{3}-1}=1, a_{5}=\frac{a_{3}+a_{4}}{a_{3} a_{4}-1}=2$. From $a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+a_{n+2}$, $a_{n+1} a_{n+2} a_{n+3}=a_{n+1}+a_{n+2}+a_{n+3}$, subtracting gives $a_{n+3}-a_{n}=a_{n+1} a_{n+2}\left(a_{n+3}-a_{n}\right), a_{n+1} a_{n+2} \neq 1$, so $a_{n+3}=a_{n}$, which means $\left\{a_{n}\right\}$ is a periodic sequence. $a_{1}+a_{2}+a_{3}=6,1999=333 \times 6+1$, then $S_{1999}=6 \times 666+1$.