1. Given non-zero real numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}$ $=1$, and $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c$. $\left(\frac{1}{a}+\frac{1}{b}\right)=-3$. Then $a+b+c=$ $\qquad$
$$
\text { Two, 1. } a+b+t=0, \pm 1 \text { . }
$$

When $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)=-3$
clear the denominators, and factorize
$$
(a+b+c)(a b+b c+c a)=0,
$$

then $a+b+c=0$ or $a b+b c+c a=0$.
$$
\begin{array}{l}
\text { Also, } a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+c a) \\
\quad=1, \\
\therefore(a+b+c)^{2}=1 \text {, so } a+b+c= \pm 1 . \\
{[\text { Oh, } a+b+c= \pm 1,0 .}
\end{array}
$$