  We consider a 3--body system in $\mathbb{R}^3$ with non--positive potentials
and non--negative essential spectrum. Under certain requirements on the fall
off of pair potentials it is proved that if at least one pair of particles has
a zero energy resonance then a square integrable zero energy ground state of
three particles does not exist. This complements the analysis in \cite{1},
where it was demonstrated that square integrable zero energy ground states are
possible given that in all two--body subsystems there is no negative energy
bound states and no zero energy resonances. As a corollary it is proved that
one can tune the coupling constants of pair potentials so that for any given
$R, \epsilon >0$: (a) the bottom of the essential spectrum is at zero; (b)
there is a negative energy ground state $\psi(\xi)$, where $\int |\psi(\xi)|^2
= 1$; (c) $\int_{|\xi| \leq R} |\psi(\xi)|^2 < \epsilon$.
