3. As shown in Figure 1, the three vertices of the equilateral $\triangle D E F$ with side length $n$ are exactly on the sides of the equilateral $\triangle A B C$ with side length $m$. Then the inradius of $\triangle A E F$ is ( ).
(A) $\frac{\sqrt{3}}{6}(m-n)$
(B) $\frac{\sqrt{3}}{4}(m-n)$
(C) $\frac{\sqrt{3}}{3}(m-n)$
(D) $\frac{\sqrt{3}}{2}(m-n)$
3. A.

It is easy to prove that $\triangle A E F, \triangle B D E, \triangle C F D$ are congruent.
Thus, $S_{\triangle A B F}=S_{\triangle B D E}=S_{\triangle C F D}, A F=B E$.
Since $S_{\triangle A B C}=\frac{\sqrt{3}}{4} m^{2}, S_{\triangle D E F}=\frac{\sqrt{3}}{4} n^{2}$, we have
$$
S_{\triangle A E F}=\frac{1}{3}\left(S_{\triangle A B C}-S_{\triangle D E F}\right)=\frac{\sqrt{3}}{12}\left(m^{2}-n^{2}\right) \text {. }
$$

Let the inradius of $\triangle A E F$ be $r$.
By the area formula of a triangle, we know
$$
\begin{array}{l}
S_{\triangle A E F}=\frac{1}{2}(A E+E F+A F) r=\frac{1}{2}(m+n) r . \\
\text { Therefore, } r=\frac{2 S_{\triangle A E F}}{A E+E F+A F}=\frac{\sqrt{3}(m-n)}{6} .
\end{array}
$$