3-Phase Generator Frame Ground

The circuit will both be at higher voltage (well the parts of it) and will self-adjust. The reasoning behind this is pretty simple: First, the phases are out of phase, or to explain it simply, their voltage peaks happen at different times, so while one is at its maximum voltage, the two other phases will be at lower voltage than the one at maximum is, so basically, the load will be at the voltage between phases. Since we don't have the neutral conductor, the current which was going through it will have to go through one of the phases and that's how the circuit will "self-adjust". You can see that from the simulation here.

When the switch is closed, all 3 loads are at the same voltage of 5 V compared to the ground. Once the switch is open, the voltages at the loads will change. So the lightest load will drop the least voltage and the greatest load will drop the most voltage.

If you have resistors from red to blue, from blue to yellow and from yellow to red (i.e. a delta formation load) then the power is the sum of the individual line voltages (squared) divided by the individual resistance across each line:

$$P = \frac{V_{RB}^2}{R_{RB}} + \frac{V_{BY}^2}{R_{BY}} + \frac{V_{YR}^2}{R_{YR}}$$

If you have resistors in a star formation this is more difficult unless you have a neutral wire commoning the three resistors. If you do then measure the individual phase voltages and do individual power calculations then add the three powers to give total load power.

If you don't have a neutral then you will have to calculate the star-point voltage relative to red, blue and yellow respectively. Then you'll have three voltages and three resistors and power is the sum of the individual powers.