Dennis,
IMHO there is no 'right answer' because you can always put more and more effort into closer and closer refinements of the answer.
Each one of the solutions provided made different approximations and simplifications, and these intentional factors are on top of any errors. Each simplification will introduce errors, and it becomes a judgement call to determine if the expected error is small enough. Remember that there are things that no-one bothers bringing in to these calculations, such as manufacturing variability or supply voltage variability. A 1200VA lamp may draw more or less than 1200VA, and a 240V supply may be more or less than 240V...and a 12ga wire may be thicker or thinner than 12ga
On top of this, different people are using different lamp loads in their examples. In the below, I will re-calculate the various approaches as best I can, using a supply voltage of 240V, a lamp load of 1200VA, and 13 lamps.
1) jbwhite (the original post) essentially was saying 'figure out the worst imbalance, and then assume a balanced three phase load where each leg has as many lamps as the most heavily loaded leg'. Pretty reasonable and simple. There was an error in the calculation, in that he calculates the single phase load but doesn't correctly take this single phase load and combine it into a three phase load. My recalculation gives a load of 43.3A/phase.
2) Dennis, you play a slight variation, figuring out the worst imbalance as seen by a single phase leg, and then assuming a balanced three phase system that looks like that. You don't bother with vector addition to get the single phase load, so you are basically doing the 'assume the system is balanced' but at the level of each leg, rather than for the whole three phase system. Still quite reasonable and simple. I get 39A/phase.
3) iwire basically says 'treat this as a balanced three phase system, it will be close enough.' My recalculation gives 37.5A/phase.
4) bob gives the first answer calling for vector addition. Smart$ later does the vector addition with a spread sheet.
5) Lou calculates it two ways. First by figuring the single phase loads, and adding these up by leg, without compensating for the phase angle difference, and then by assuming a worst case balanced three phase load.
For Lou's first method I recalculate 45A on the heavily loaded legs. Note: 39 * (2/1.732) = 45; so the difference between Lou's and Dennis' method is in how loads on individual branches were added up.
For Lou's second method, I get 43.3A/phase, the same as the original method with the 1.732 factor brought in.
6) bob gives an approach that drops the vector addition: calculate this by assuming a balanced three phase load for the _balanced_ portion, and then simply adding the unbalanced portion, without worrying about vectors. 12 lamps balanced gives 34.7A. The unbalanced lamp is drawing 5A, so I just add this to the legs in question, giving 39.7A.
7) Smart$ presents a spreadsheet that does the calculations using a simplified equation that would give the vector addition results if you assume an exact 120 degree phase angle difference. This is itself an approximation, but far closer than simply adding the values up. My recalculation gives 39A on the heavily loaded legs (I get a different number because Smart$ assumed 1200W and 0.9 power factor; I am assuming 1200VA with the power factor in the 1200 number)
IMHO approach 6 is the one that makes the most sense if doing this on paper and pencil: figure out the balanced load, and then add something for the unbalanced portion. It isn't as accurate as the spreadsheet approach, but looks 'close enough' to my eye; the real question being: are there situations where you could save on wire or OCPD if you used the spreadsheet rather than approach 6? Probably; if approach 6 gives a result of 40.6A I would be inclined to take a closer look at the results.
Like most calculations, you can do a 'rough guess' first, and if the results are below a threshold, then simply decide that no further calculation is necessary; if you use a rough and ready VD calculation that you know always gives high results, and you get 2%, then you don't need to do the accurate calculation to determine that you actually expect a voltage drop of 1.673%
-Jon
