  The hidden charm $X(3872)$ resonance is usually thought to be a $D^{*0}
\bar{D}^0$ meson-antimeson molecule with quantum numbers $J^{PC} = 1^{++}$. If
this is the case, there is the possibility that there might be three body bound
states with two charmed mesons and a charmed antimeson. Here we argue that the
theoretical existence of this type of three body molecules is expected from
heavy quark spin symmetry. If applied to the two body sector, this symmetry
implies that the interaction of the $D^{*0} \bar{D}^{*0}$ meson-antimeson pair
in the $J^{PC} = 2^{++}$ channel is the same as in the $J^{PC} = 1^{++}$
$D^{*0} \bar{D}^0$ case. From this we can infer that the $J^P = 3^{-}$ $D^{*0}
D^{*0} \bar{D}^{*0}$ molecule will be able to display the Efimov effect if the
scattering length of the $2^{++}$ channel is close enough to the unitary limit.
Heavy quark spin symmetry also indicates that the $J^P = 2^{-}$ $D^{*0} D^{*0}
\bar{D}^0$ molecule is analogous to the $J^P = 3^{-}$ $D^{*0} D^{*0}
\bar{D}^{*0}$ one. That is, it can also have a geometric spectrum. If we
consider these triply heavy trimers in the isospin symmetric limit, the Efimov
effect disappears and we can in principle predict the fundamental state of the
$2^-$ $D^*D^*\bar{D}$ and $3^-$ $D^* D^* \bar{D}^*$ systems. The same applies
to the $B^*B^* \bar{B}^*$ system: if the $Z_b(10650)$ is an isovector $B^*
\bar{B}^*$ molecule then the $0^-$ isodoublet and the $1^-$, $2^-$ isoquartet
$B^* B^* \bar{B}^*$ trimers might bind, but do not display Efimov physics.
Finally from heavy flavour symmetry it can be argued that scattering in the $B
D$ two-body system might be resonant. This would in turn imply the possibility
of Efimov physics in the $B B D$ three body system.
