3. As shown in Figure 2, in a regular tetrahedron $A B C D$ with edge length 1, $M$ and $N$ are the midpoints of $A D$ and $B C$, respectively. The minimum area of the section cut by the plane through $M N$ in this tetrahedron is ( ).
(A) $\frac{1}{4}$
(B) $\frac{1}{12}$
(C) $\frac{\sqrt{2}}{4}$
(D) $\frac{\sqrt{3}}{8}$
3. A.

First, the plane through $M N$ must intersect a pair of edges at one point each, and we should set it to intersect $A B$ at $E$ and $C D$ at $F$, as shown in Figure 5.
Notice that
$$
S_{\triangle E M V}=\frac{1}{2} M N \cdot h_{E-M V}
$$

and $h_{\kappa-u v} \geqslant h_{A H-u v}$. Since $A B$ and $M N$ are common perpendicular lines, then $h_{A R-M V}=O E_{1}=\frac{\sqrt{2}}{4}$.
Therefore, $S_{\triangle B M} \geqslant \frac{1}{2} M N \cdot O E_{1}=\frac{1}{8}$.
Similarly, $S_{\triangle Y K V} \geqslant \frac{1}{8}$.
Thus, $S_{\text {quadrilateral:MF: }} \geqslant \frac{1}{4}$.
The equality holds when $E$ and $F$ are the midpoints of $A B$ and $C D$ respectively. At this time, $E, M, F, N$ are coplanar, which meets the conditions.