3. If $\frac{a^{2}}{b+c-a}+\frac{b^{2}}{c+a-b}+\frac{c^{2}}{a+b-c}=0$, then $\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}=$ $\qquad$
3.1 or $-\frac{3}{2}$.

Let $\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}=k$.
Then, by the given condition,
$$
\begin{array}{l}
a\left(k-\frac{b}{c+a-b}-\frac{c}{a+b-c}+\frac{a}{b+c-a}\right)+ \\
b\left(k-\frac{c}{a+b-c}-\frac{a}{b+c-a}+\frac{b}{c+a-b}\right)+ \\
c\left(k-\frac{a}{b+c-a}-\frac{b}{c+a-b}+\frac{c}{a+b-c}\right)=0 \\
\Rightarrow(a+b+c) k-\frac{a(b+c-a)}{b+c-a}- \\
\quad \frac{b(c+a-b)}{c+a-b}-\frac{c(a+b-c)}{a+b-c}=0 \\
\Rightarrow(a+b+c)(k-1)=0 .
\end{array}
$$

Thus, $a+b+c=0$ or $k-1=0$.
When $a+b+c=0$,
$$
b+c=-a, c+a=-b, a+b=-c \text {, }
$$

at this time, $k=\frac{a}{-2 a}+\frac{b}{-2 b}+\frac{c}{-2 c}=-\frac{3}{2}$.
In conclusion, $k=1$ or $-\frac{3}{2}$.