1. Given positive integers $a, b, c$ satisfy
$$
\left\{\begin{array}{l}
a b+b c+c a+2(a+b+c)=8045, \\
a b c-a-b-c=-2 .
\end{array}\right.
$$

then $a+b+c=$ $\qquad$
$$
-1.2012 .
$$

Note that
$$
\begin{array}{l}
(a+1)(b+1)(c+1) \\
=a b c+a b+b c+c a+a+b+c+1 \\
=8045+(-2)+1=8044 .
\end{array}
$$

Since $a, b, c$ are positive integers, we have
$$
a+1 \geqslant 2, b+1 \geqslant 2, c+1 \geqslant 2 \text{. }
$$

Therefore, 8044 can only be factored as $2 \times 2 \times 2011$. Hence $a+b+c=2010+1+1=2012$.