3. Given $a^{2}+b^{2}+c^{2}=1$. Then the range of $a b+b c+a c$ is . $\qquad$
3. $\left[-\frac{1}{2}, 1\right]$.

Obviously, $a b+b c+a c \leqslant a^{2}+b^{2}+c^{2}=1$.
On the other hand,
$$
\begin{array}{l}
a b+b c+a c=a(b+c)+b c \\
\geqslant-\frac{a^{2}+(b+c)^{2}}{2}+b c \\
=-\frac{a^{2}+b^{2}+c^{2}}{2}=-\frac{1}{2},
\end{array}
$$

Equality holds only when $a+b+c=0$, and $a^{2}+b^{2}+c^{2}=1$.
Therefore, the range is $\left[-\frac{1}{2}, 1\right]$.