Electronic – How is Line and Phase Voltage the same for a Delta connected system

Your question misses the point.

As there is a neutral conductor between the generator and the load, any imbalance between currents in the three phases is taken up by current in the neutral conductor.

Balance, when feeding multiple single phase loads, is purely statistical. If you power enough houses, then you will find that the current flow in neutral is noisy, generally small with the occasional peak.

Balance, when driving 3 phase loads, is part of the specification of the load. 3 phase motors and heaters will usually have a spec for what balance they are intended to achieve.

The connection of neutral to local ground is a different issue. If the whole system was connected to ground at a single point, say at the source, then of course there would be no current flow through that tie conductor, and no voltage difference between those points. However, the neutral conductor arriving at a distant point would have a voltage drop across it due to the current flowing, and a potential maybe a few volts from ground.

If there is a further local tie point, then that potential of a few volts will drive a current through that tie point. What is happening now is that the neutral current is being shared between the neutral conductor, and the ground, between source and load.

Generally, injecting significant currents into ground is deemed to be a BAD THING, and the many regulations around how to load 3 phase systems, how to size neutral conductors, when and where you can and can't tie neutral to local ground, are in pursuit of minimising ground currents.

It comes down to the fact that in an ideal system, each phase has a sine waveform, and the well known relationship

$$sin(x)^2 + cos(x)^2 = 1 $$ for all X, and power is proportional to V^2.

Now that relationship obviously applies to 2 phases 90 degrees apart. How can you apply it to 3 phases 120 degrees apart?

Take the first phase as Sin(X). The other two are $$sin(x + 120 degrees)$$ and $$sin(x - 120 degrees)$$

A bit of vector analysis (draw the three out to illustrate) lets you resolve each of the latter two phases into sine and cosine components. So $$sin(x + 120) = A * sin(x) + B * cos(x)$$ I'll let you figure out the values of A and B.

Repeat for the other phase $$sin(x + 120) = C * sin(x) + D * cos(x)$$

Now square each of the three phases and sum over the squares... this is a bit of an exercise, but you should get

$$sin(x)^2 + cos(x)^2$$

which is of course 1.