3. Let the complex numbers be
$$
z_{1}=\sin \alpha+2 \mathrm{i}, z_{2}=1+\cos \alpha \cdot \mathrm{i}(\alpha \in \mathbf{R}) \text {. }
$$

Then the minimum value of $f=\frac{13-\left|z_{1}+\mathrm{i} z_{2}\right|^{2}}{\left|z_{1}-\mathrm{i} z_{2}\right|}$ is . $\qquad$
3. 2 .
$$
\begin{array}{l}
\text { Let }\left|z_{1}-\mathrm{i} z_{2}\right|=\sqrt{(\sin \alpha+\cos \alpha)^{2}+1^{2}} \\
=\sqrt{2+\sin 2 \alpha}=t .
\end{array}
$$

Then $t \in[1, \sqrt{3}]$, and at this time we have
$$
\begin{array}{l}
\left|z_{1}+\mathrm{i} z_{2}\right|^{2}=(\sin \alpha-\cos \alpha)^{2}+3^{2} \\
=10-\sin 2 \alpha=12-t^{2} . \\
\text { Hence } f=\frac{13-\left|z_{1}+\mathrm{i} z_{2}\right|^{2}}{\left|z_{1}-\mathrm{i} z_{2}\right|}=\frac{1+t^{2}}{t} \geqslant 2 .
\end{array}
$$

When $t=1$, i.e., $\alpha=k \pi-\frac{\pi}{4}(k \in \mathbf{Z})$, the minimum value of $f$ is 2.