3. In a regular tetrahedron $ABCD$, $AC=1$, $M$ is the midpoint of $AC$, $N$ is the centroid of $\triangle BCD$, and $DE \perp AB$, with the foot of the perpendicular being $E$. Then the angle formed by the skew lines $MN$ and $DE$ is
A. $90^{\circ}$
B. $\arccos \frac{5}{6 \sqrt{3}}$
C. $\arccos \frac{1}{\sqrt{3}}$
D. $\arccos \frac{5}{6}$
3. $\mathrm{B}$ As shown in the figure, let $F$ be the midpoint of $CD$.

Taking $E$ as the origin, $EF$ and $EA$ as the $y$-axis and $z$-axis respectively to establish a rectangular coordinate system.
Then $A\left(0,0, \frac{1}{2}\right), B\left(0,0,-\frac{1}{2}\right)$,
Since $CD \perp$ plane $yOz$, $F$ is the midpoint of $CD$, $EF=\sqrt{DE^{2}-DF^{2}}=\frac{\sqrt{2}}{2}$,
Thus, $D\left(-\frac{1}{2}, \frac{\sqrt{2}}{2}, 0\right), C\left(\frac{1}{2}, \frac{\sqrt{2}}{2}, 0\right)$. $M$ is the midpoint of $AC$, so $M\left(\frac{1}{4}, \frac{\sqrt{2}}{4}, \frac{1}{4}\right)$.
The coordinates of $N$ are $\frac{1}{3}\left[\left(0,0,-\frac{1}{2}\right)+\left(\frac{1}{2}, \frac{\sqrt{2}}{2}, 0\right)+\left(-\frac{1}{2}, \frac{\sqrt{2}}{2}, 0\right)\right]=\left(0, \frac{\sqrt{2}}{3},-\frac{1}{6}\right)$, so $\overrightarrow{MN}=\left(-\frac{1}{4}, \frac{\sqrt{2}}{12},-\frac{5}{12}\right)$,
Therefore, $\overrightarrow{MN} \cdot \overrightarrow{ED}=\left(-\frac{1}{4}, \frac{\sqrt{2}}{12},-\frac{5}{12}\right) \cdot\left(-\frac{1}{2}, \frac{\sqrt{2}}{2}, 0\right)=\frac{5}{24}$,
$$
\theta=\arccos \frac{\overrightarrow{MN} \cdot \overrightarrow{ED}}{|\overrightarrow{MN}| \cdot|\overrightarrow{ED}|}=\arccos \frac{\frac{5}{24}}{\sqrt{\frac{3}{4}} \cdot \frac{1}{12} \sqrt{9+2+25}}=\arccos \frac{5}{6 \sqrt{3}}
$$