I understand that the nitrogen not attached to the hydrogen in Imidazole is the most basic of the 3 because of the lone pair that is in the same plane of the aromatic system.
What you say is true, but I don't think it explains why it is the most basic N. The N in imidazole (not the NH) is sp
2 hybridised (and not delocalised) and in pyrrolidine the NH is sp
3 hybridised. Typically, sp
3 lone pairs (comparing the same atom) are more basic than sp
2 due to the larger size of the sp
3 orbital - this is not observed in this example, so there must be something else going on. That something else is resonance. Protonation of the N (not NH) in imidazole gives a resonance stabilised cation in which the positive charge is shared equally over the two Ns - a cation more stable than the localised protonated pyrrolidine.
For the NH in imidazole vs NH in pyrrolidine, think about the relative availability of the lone pairs for picking up a proton. Resonance is the key once more.