3. Let $O$ be a point inside $\triangle A B C$, and let $D$, $E$, $F$ be the projections of $O$ onto $B C$, $C A$, $A B$ respectively. When $\frac{B C}{O D}+\frac{C A}{O E}+\frac{A B}{O F}$ is minimized, $O$ is the ( ) of $\triangle A B C$.
(A) Centroid
(B) Centroid
(C) Incenter
(D) Circumcenter
3.C.

Let the area and perimeter of $\triangle A B C$ be denoted as $S$ and $l$, respectively. By Cauchy's inequality, we have
$$
\begin{array}{l}
\left(\frac{B C}{O D}+\frac{C A}{O E}+\frac{A B}{O F}\right)(B C \cdot O D+C A \cdot O E+A B \cdot O F) \\
\geqslant(B C+C A+A B)^{2}=l^{2} . \\
\text { Also, } B C \cdot O D+C A \cdot O E+A B \cdot O F=2 S,
\end{array}
$$

From (1), we get
$$
\frac{B C}{O D}+\frac{C A}{O E}+\frac{A B}{O F}>\frac{l^{2}}{2} \text {. }
$$

Since $\frac{l^{2}}{2 S}$ is a constant, the minimum value is achieved when the equality in (2) holds. When the equality in (2) holds, the equality in (1) also holds. This occurs if and only if $\sqrt{\frac{B C}{O D}}: \sqrt{B C \cdot O D}=\sqrt{\frac{C A}{O E}}: \sqrt{C A \cdot O E}=\sqrt{\frac{A B}{O F}}: \sqrt{A B \cdot O F}$, i.e., when $O D=O E=O F$. Therefore, $O$ is the incenter of $\triangle A B C$.