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 $E$, $F$, and $B_{1}$ is ( ${ }^{\prime}$ ).
(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 the figure, $B_{1} F=\frac{\sqrt{5}}{2} a, M N \underline{\mathbb{L}} B F, C M=\frac{1}{2}$. $A_{1} F=\frac{1}{4} a, C_{1} M=\frac{3}{4} a$, so, $M B_{1}=\frac{5}{4} a$. Connect $M F$, let $\angle M B_{1} F=\theta$,
$$
\begin{array}{l}
M F \\
\sqrt{(\sqrt{2} a)^{2}+\left(\frac{1}{4} a\right)^{2}} \\
=\frac{\sqrt{33}}{4} a .
\end{array}
$$

In $\triangle F B_{1} M$, by the cosine rule, we have
$$
\cos \theta=\frac{B_{1} F^{2}+B_{1} M^{2}-F M^{2}}{2 B_{1} F \cdot B_{1} M}=\frac{3 \sqrt{5}}{25} .
$$

Then $\sin \theta=\frac{2 \sqrt{145}}{25}$.
Thus, $S_{\text {OB } B_{1} M N F}=B_{1} F \cdot B_{1} M \cdot \sin \theta=\frac{\sqrt{29}}{4} a^{2}$,
Notice that $S_{\triangle N E X}=\frac{1}{2} \times N E \cdot N K \cdot \sin \theta$
$$
\begin{array}{l}
\quad=\frac{1}{2} \times\left(\frac{1}{2} M N\right) \times\left(\frac{1}{3} N F\right) \cdot \sin \theta \\
=\frac{1}{12} M N \cdot N F \cdot \sin \theta \\
=\frac{1}{12} S_{\square B_{1} M N F},
\end{array}
$$

It is clear that $S_{\text {millipede }}=\frac{11}{12} S_{\square B_{1} M N F}$
$$
=\frac{11}{12} \times \frac{\sqrt{29}}{4} a^{2}=\frac{11 \sqrt{29}}{48} a^{2} .
$$