This triangle is equilateral.

since $0<a,b,c<\pi$ we'll get that $\sin(\alpha),\sin(\beta),\sin(\gamma)\in\mathbb{R}^{+}$ from $AM\geq GM;~(\sin(\alpha)+\sin(\beta)+\sin(\gamma))(\sin^2(\alpha)+\sin^2(\beta)+\sin^2(\gamma))$ $\geq (3\sqrt[3]{\sin(\alpha)\sin(\beta)\sin(\gamma)})(3\sqrt[3]{\sin^2(\alpha)\sin^2(\beta)\sin^2(\gamma)})$ $=9\sin(\alpha)\sin(\beta)\sin(\gamma)~~~~~(*)$ which equality case holds at $\alpha=\beta=\gamma=\frac{\pi}{3}$ from well known trigonometric identity $a\cos\alpha+b\cos\beta+c\cos\gamma=4R\sin(\alpha)\sin(\beta)\sin(\gamma)$ and $a\sin\alpha+b\sin\beta+c\sin\gamma=2R(\sin^2(\alpha)+\sin^2(\beta)+\sin^2(\gamma))$ from law of sine; $\frac{2p}{R}=2(\sin(\alpha)+\sin(\beta)+\sin(\gamma))$ since $\frac{a\cos\alpha+b\cos\beta+c\cos\gamma}{a\sin\alpha+b\sin\beta+c\sin\gamma}=\frac{2p}{9R}$ $\Rightarrow \frac{4R\sin(\alpha)\sin(\beta)\sin(\gamma)}{2R(\sin^2(\alpha)+\sin^2(\beta)+\sin^2(\gamma))}=\frac{2(\sin(\alpha)+\sin(\beta)+\sin(\gamma))}{9}$ $\Rightarrow (\sin(\alpha)+\sin(\beta)+\sin(\gamma))(\sin^2(\alpha)+\sin^2(\beta)+\sin^2(\gamma))=9\sin(\alpha)\sin(\beta)\sin(\gamma)$ from $(*)$ we can conclude that $\alpha=\beta=\gamma=\frac{\pi}{3}$ $\therefore~\frac{a\cos\alpha+b\cos\beta+c\cos\gamma}{a\sin\alpha+b\sin\beta+c\sin\gamma}=\frac{2p}{9R}\Leftrightarrow \triangle ABC$ is a equilateral triangle