For a more complicated but more conservative VA accounting method, keep 6 separate VA totals, one each for A, B, and C, and one each for AB, BC, and CA. Then

1) Each L-N load VA gets assigned to A, B, or C accordingly

2) Each L-L load VA gets assigned to AB, BC, or CA accordingly

3) For balanced 3 phase loads L-L-L, assign 1/3 of the VA to each of A, B, and C.

Now at the end you have 3 VA totals for AB, BC, and CA. Take the smallest of those, and subtract that value from each of the AB, BC, and CA totals and add it to each of the A, B, and C totals. This is just moving VA around in the accounting, and it represents the possibly balanced portion of the L-L loads.

That leaves you with one of AB, BC, and CA equal to 0 VA. For ease of presentation I'll assume it's CA; the other cases are comparable. The line currents are now approximated by:

I_{A} = VA_{A} / 120V + VA_{AB} / 208V

I_{B} = VA_{B} / 120V + min(VA_{AB},VA_{BC}) / 120V + |VA_{AB}-VA_{BC}| / 208V

I_{C} = VA_{C} / 120V + VA_{BC} / 208V

This will be conservative (any error is on the high side) if the L-L loads all have the same power factor. In which case min(VA_{AB},VA_{BC}) represents the load on leg B where current from AB and BC partially cancel, and |VA_{AB}-VA_{BC}| represents the rest of the load from the higher of the two values.

Cheers, Wayne