28. Let $A \geqslant 60^{\circ}$, then $B+C \leqslant 180^{\circ}-60^{\circ}=120^{\circ}$. $\sin 3 A+\sin 3 B+\sin 3 C=\sin 3 A+2 \sin \frac{3}{2}(B+C) \cos \frac{3}{2}(B-C) \leqslant \sin 3 A+2 \sin \frac{3}{2}(B+C)$. Let $\alpha=\frac{3}{2}(B+C)$, then $0 \leqslant \alpha \leqslant 180^{\circ}$, and $A=180^{\circ}-(B+C)=180^{\circ}-\frac{2}{3} \alpha$. Therefore, $\sin 3 A+\sin 3 B+\sin 3 C \leqslant \sin \left(3 \times 180^{\circ}-2 \alpha\right)+2 \sin \alpha=\sin 2 \alpha+2 \sin \alpha=2 \sin \alpha(1+\cos \alpha)=8 \sin \frac{\alpha}{2} \cos ^{3} \frac{\alpha}{2}$. By the AM-GM inequality, we get $\sin \frac{\alpha}{2} \cos ^{3} \frac{\alpha}{2}=\sqrt{\sin ^{2} \frac{\alpha}{2} \cos ^{6} \frac{\alpha}{2}}=\sqrt{\frac{1}{3} \cdot 3 \sin ^{2} \frac{\alpha}{2} \cos ^{6} \frac{\alpha}{2}} \leqslant \sqrt{\frac{1}{3}\left[\frac{3 \sin ^{2} \frac{\alpha}{2}+\cos ^{2} \frac{\alpha}{2}+\cos ^{2} \frac{\alpha}{2}+\cos ^{2} \frac{\alpha}{2}}{4}\right]^{4}} \leqslant \frac{3 \sqrt{3}}{16}$. Therefore, $\sin 3 A+\sin 3 B+\sin 3 C \leqslant \frac{3}{2} \sqrt{3}$, and equality holds if and only if $A=140^{\circ}, B=C=20^{\circ}$.