Well I've always wondered, and my Kaplan (page 512) says that the binding energy of 3H3 is greater (8.482mev) than that of 3He3 (7.711Mev) by (0.711Mev). And it attributes the lesser binding energy to the coulomb repulsion. And that would indicate the 3He3 nuclide to be the larger. And since a deuteron would repel a proton with a greater force than a neutron, it's more logical and more probable to accumulate an additional neutron, and the question then becomes as to why 1 of 2 nearby neutrons would want to become a proton.