(3) For the tetrahedron $A-BCD$, the projection of vertex $A$ onto the base $BCD$ is point $O$, and the distances from point $O$ to the three side faces are equal. Then point $O$ must be the ( ) of $\triangle BCD$.
(A) Centroid
(B) Orthocenter
(C) Incenter
(D) Circumcenter
(3) Let $OE \perp$ plane $ABC$ at point $E$, $OF \perp$ plane $ACD$ at point $F$. Extend $AE$ and $AF$ to intersect $BC$ and $CD$ at points $G$ and $H$, respectively, and connect $OG$ and $OH$. In right triangles $\triangle OEA$ and $\triangle OFA$, $OE=OF$, $OA=OA$, so right triangles $\triangle OEA \cong \mathrm{Rt} \triangle OFA$, which means $\angle OAE=\angle OAF$. Therefore, right triangles $\triangle OAG \cong \mathrm{Rt} \triangle OAH$, and $OG=OH$. Since $OE \perp BC$, $OF \perp CD$, and $AO \perp$ plane $BCD$, it follows that $BC \perp$ plane $AOG$, $CD \perp$ plane $AOH$, so $BC \perp OG$, $CD \perp OH$. Thus, $O$ is the incenter of $\triangle BCD$. Therefore, the correct choice is $\mathrm{C}$.