3. As shown in the figure, in the regular tetrahedron $ABCD$, $E$ is on edge $AB$, and $F$ is on edge $CD$, such that $\frac{AE}{EB} = \frac{CF}{FD} = \lambda (0 < \lambda < +\infty)$. Let $f(\lambda) = \alpha_{\lambda} + \beta_{\lambda}$, where $\alpha_{\lambda}$ is the angle between $EF$ and $AC$, and $\beta_{\lambda}$ is the angle between $EF$ and $BD$. Then,
A. $f(\lambda)$ is monotonically increasing on $(0, +\infty)$
B. $f(\lambda)$ is monotonically decreasing on $(0, +\infty)$
C. $f(\lambda)$ is monotonically increasing on $(0, 1)$ and monotonically decreasing on $(1, +\infty)$
D. $f(\lambda)$ is a constant on $(0, +\infty)$
3. D Since $A B C D$ is a regular tetrahedron, $A C \perp B D$. Draw $E G / / A C$ intersecting $B C$ at $G$, and connect $G F$, then $\alpha_{\lambda}=\angle G E F$, and $\frac{O G}{G B}=\frac{A E}{E B}=\frac{C F}{F D}$, so $G F / / B D$. Therefore, $G F \perp G E$, and $\beta_{\lambda}=\angle E F G$. Thus, $f(\lambda)=\alpha_{\lambda}+\beta_{\lambda}=90^{\circ}$ is a constant.