12. In the arithmetic sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{20} \simeq \frac{1}{a}, a_{201}=\frac{1}{b}, a_{2012}=\frac{1}{c} \text {. }
$$

Then $1992 a c-1811 b c-181 a b=$
12. 0 .

Let the common difference of the arithmetic sequence be $d$. Then, according to the problem, we have
$$
\begin{array}{l}
a_{201}-a_{20}=\frac{a-b}{a b}=181 d, \\
a_{2012}-a_{201}=\frac{b-c}{b c}=1811 d, \\
a_{2012}-a_{20}=\frac{a-c}{a c}=1992 d .
\end{array}
$$

Therefore, $1992 a c-1811 b c-181 a b$
$$
=\frac{a-c}{d}+\frac{c-b}{d}+\frac{b-a}{d}=0 .
$$