2. Given $a^{2}+b^{2}+c^{2}=1$,
$$
a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)=-3 \text {. }
$$

Find the value of $a+b+c$.
$$
\begin{array}{l}
\text { From the given equation, we have } \\
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0 \\
\Rightarrow a+b+c=0 \text { or } \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0 .
\end{array}
$$

From the latter, we get $a b+b c+c a=0$.
$$
\begin{aligned}
\text { Therefore, }(a+b+c)^{2} & =a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=1 \\
\Rightarrow a+b+c & = \pm 1 .
\end{aligned}
$$

In summary, $a+b+c=0,-1,1$.