3. D.

As shown in Figure 2, it is easy to see that points $B, C, E, F$ are concyclic. Therefore, $\triangle A E F \backsim \triangle A B C$. Thus, $\cos A=\frac{A F}{A C}=\frac{E F}{B C}=\frac{5}{13}$, $\sin A=\sqrt{1-\cos ^{2} A}=\frac{12}{13}$. Hence, $B E=A B \sin A=26 \times \frac{12}{13}=24$.
3. Given $A D, B E, C F$ are the altitudes of the acute $\triangle A B C$. If $A B=26, \frac{E F}{B C}=\frac{5}{13}$, then the length of $B E$ is ( ).
(A) 10
(B) 12
(C) 13
(D) 24