The neutral conductor will only carry the third-harmonic currents. This is because:
Positive-sequence (and negative-sequence) balanced currents have the same magnitude but are displaced by 120 degrees from each other. Due to the 120-degree displacement, they add up to 0 at the neutral connection point.
1<0 + 1<-120 + 1<+120 = 0
48<0 + 48<-120 + 48<+120 = 0
Zero-sequence currents have both the same magnitude and same phase angles, so they are additive at the neutral connection point. So in this problem: