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The sources are all the same and the three phase loads are all the same, exactly, so you can handle this as a completely balanced load. This means you can calculate the power in one phase and multiply by 3 to get the total.
The lowest most series circuit made from the 1 ohm resistor and j2 ohm inductor can be ignored completely because since this is a balanced load not only is the junction of all the sources equal to zero but the junction of all the loads (at the point where all the j12 inductors connect together) is also zero so there is no current flow in that lowest circuit in the drawing.
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