Motivated by previous works in which the $\rho B^{\ast}$ and the $B^{\ast} \bar{B}^{\ast}$ interactions were studied, we calculate the interaction of the three-body system $\rho B^{\ast} \bar{B}^{\ast}$. We know that the $\rho B^{\ast}$ interaction in $J=2$ is stronger than in any other possible spin, and also an attractive interaction was found in the $B^{\ast} \bar{B}^{\ast}$ system, producing in both cases a bound state. Then we search for a three-body bound state in the $\rho B^{\ast} \bar{B}^{\ast}$ system assuming that the $B^{\ast} \bar{B}^{\ast}$ is forming a cluster and letting the lighter $\rho$ meson interact with the $B$-mesons always in a spin two configuration. This fact justifies the using of the Fixed Center approximation, and considering the $J=2$ $\rho B^{\ast}$ interaction we find a $J=3$ three body meson molecule solving the Faddeev equations. As a consequence of the strongly attractive two-body interaction in the different subsystems , a $J=3$ three-body state is found, providing a prediction of an exotic state.

Based on "States of $\rho B^{\ast} \bar{B}^{\ast}$ with $J=3$ within the Fixed Center Approximation to Faddeev equations" Eur. Phys. J. A in print. arXiv:1510.06570 [hep-ph].