1. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a+b+c=1, \\
\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}=1 .
\end{array}
$$

Then $a b c=$ . $\qquad$
$$
\begin{array}{l}
\frac{1}{1-2 c}+\frac{1}{1-2 a}+\frac{1}{1-2 b}=1 \\
\Rightarrow(1-2 a)(1-2 b)+(1-2 b)(1-2 c)+(1-2 a)(1-2 c) \\
\quad=(1-2 a)(1-2 b)(1-2 c) \\
\Rightarrow 2-2(a+b+c)=8 a b c \\
\Rightarrow a b c=0 .
\end{array}
$$