3. Given that point $P$ is inside $\triangle A B C$, and perpendiculars from $P$ to $B C$, $C A$, and $A B$ intersect at points $D$, $E$, and $F$ respectively, find all points $P$ such that $\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}$ is minimized.
3. Let the area of $\triangle A B C$ be $S$, then $S=\frac{1}{2} B C \cdot P D+\frac{1}{2} C A \cdot P E+\frac{1}{2} A B \cdot F P$. By the Cauchy-Schwarz inequality, we have
$$
(B C+C A+A B)^{2} \leqslant\left(\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}\right)(B C \cdot P D+C A \cdot P E+A B \cdot F P) \text {, }
$$

Thus, $\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F} \geqslant \frac{(B C+C A+A B)^{2}}{2 S}$, and when $P$ is the incenter of $\triangle A B C$, the value of $\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}$ is minimized.