Normally in a balanced 3-phase system, the neutral conductor carries a sum total of 0 current because of how the phasors work.
I<0 + I<-120 + I<+120 = 0
But when there are zero-sequence harmonic currents, they add up in the neutral connection point and thus add up in the neutral conductor. This is because the zero-sequence currents now have the same phase angle.
So in this problem,
The fundamentals:
60<0 + 60<-120 + 60<+120 = 0
The neutral current:
3<0 + 3<0 + 3<0 = 9<0 A
NOTE: I am simply and arbitrarily using a reference angle of 0 degree for Phase A. This logic and concept still applies for any other reference angle.
Basically, only the sum of the zero-sequence currents show up in the neutral conductor.
