The short answer to your question is that “Path A” is not involved at all. Current does not flow into the ground (i.e., planet Earth) unless there is a fault somewhere in the system.
However, there are several things wrong with your sketch. First and foremost, you are incorrectly showing a connection between the current-carrying portion of the load and the ground. The ground wire at the load end is attached to the external metal portions of the load (e.g., the metal frame of a motor), not to any of the wires inside the component. If you disconnect the green wire from the load to the ground on the right side of your sketch, you will see that “Path A” does not exist anymore, as should be the case.
Secondly, when current leaves the transformer along one of the phase conductors, it will return to the transformer on one or both of the other phase conductors. At any moment in time, the total amount of current flowing in the four wires (A, B, C, and N) is zero. But the way you are showing current flows, you have 350 amps flowing away from the transformer on A, B, and C, and you have 50 amps flowing back to the transformer on N. That is not how current flows.
Finally, you can’t add current in the three phases as though you were adding up the number of apples in one box to the number of apples in another box to come up with the total number of apples. The three currents do not reach their respective peak values at the same time, so you can’t add them up as though they were all the same type of quantity. To get the numerical value of the neutral current, therefore, you don’t add up A plus B plus C and assign whatever is leftover to N. This is the formula (with the symbology of A*2 meaning A squared):
N*2 = A*2 + B*2 + C*2 –AB – AC – BC.
Charles E. Beck, P.E., Seattle
Comments based on 2008 NEC unless otherwise noted.