Why does line current equals the total apparent power divided by 3 times the phase voltage?
If your network is strong which mean that your short-circuitpower is higher than the power demand, the voltage at the input will not drop under load, than the power of the loads add up. Therefore the total power at the input is 

$$ S_{total} = S_{Z}+S_{c}+S_{equil} $$

The basic equation of apparent power for a generic device with no harmonics:

$$ S = 3 V_{phase} I $$

Therefore, in your case

$$ I = \frac{S_{total}}{3V_{phase}} $$