3. For the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $a$, $E$ is the midpoint of $C D$, and $F$ is the midpoint of $A A_{1}$. The area of the section through points $E$, $F$, and $B_{1}$ is ( ).
(A) $\frac{55 \sqrt{5}}{192} a^{2}$
(B) $\frac{\sqrt{29}}{4} a^{2}$
(C) $\frac{11 \sqrt{29}}{48} a^{2}$
(D) $\frac{7 \sqrt{29}}{8} a^{2}$
3. C.

As shown in Figure 1, draw $H G / / F B_{1}$, intersecting $D D_{1}$ and $C C_{1}$ at points $H$ and $G$ respectively, and $H F$ intersects $A D$ at point $I$. Then the pentagon $E I F B_{1} G$ is the desired cross-section.
It is easy to see that $C_{1} G=\frac{3}{4} a$,
$$
D H=\frac{1}{4} a \text {. }
$$

After calculation, we get
$$
\begin{array}{l}
B_{1} G=\frac{5}{4} a, B_{1} F=\frac{\sqrt{5}}{2} a, G F=\frac{\sqrt{33}}{4} a, \\
\cos \angle G B_{1} F=\frac{3 \sqrt{5}}{25}, \sin \angle G B_{1} F=\frac{2 \sqrt{145}}{25} .
\end{array}
$$

Therefore, $S_{\text {aHFB, }} c=\frac{5}{4} a \times \frac{\sqrt{5}}{2} a \times \frac{2 \sqrt{145}}{25}=\frac{\sqrt{29}}{4} a^{2}$.
And $S_{\triangle U H K}=\frac{1}{2} I H \cdot H E \cdot \frac{2 \sqrt{145}}{25}$
$$
\begin{array}{l}
=\frac{1}{2} \times \frac{1}{3} B_{1} G \times \frac{1}{2} H G \times \frac{2 \sqrt{145}}{25} \\
=\frac{1}{12} S_{\square \| 1 B_{1} c}=\frac{\sqrt{29}}{48} a^{2},
\end{array}
$$

Therefore, the area of the desired cross-section is $\frac{11 \sqrt{29}}{48} a^{2}$.