Reactive power overcompensation causing voltage increase at generator buses?
You are assuming lagging means less voltage.  If leading reactive power = lagging reactive power, all power would go to load, pf=1 and unity voltage.

You need 125MW + 150MW + 150MW = 425MW of real power.  You produce 182MW + 163MW + 85MW = 430MW of real power.

You need 100MVAR + 80MVAR + 85MVAR = 265MVAR of lagging reactive power. Your capacitive power correction provides 84.2MVAR + 61.2MVAR + 60.4MVAR = 205.8MVAR.  You are 59.2MVAR short.

Your over-excited generator on bus 3 provides 9MVAR (an assumption on my part due to the negative), which makes you 50.2VAR short.

So generators provide 425MW and 50.2VAR, which means 427.95MVA and a pf = 0.993.  425MW goes to loads and 50.2VAR goes back and forth between loads and generators. 
 Which is as far as my knowledge goes.

I'd guess if you provide 265MVAR - 9MVAR (bus 3 generator) = 256MVAR of capacitive power, you'd get your unity voltage.



When pf = 1, S = P and \$Q_C = Q_L\$.  The voltage of the source drives the real power load.   Your unity voltage.  Leading reactive power of the capacitors supplies the lagging reactive power of inductors (motors).

When the pf &lt; 1 and \$Q_C &lt; Q_L\$, the load is inductive.  Again, the capacitors supply reactive power to the loads, but there is not enough.  The source must supply P and \$Q_{NET} = Q_L - Q_C\$.  The source voltage, which cannot change, is providing power to drive the MW and some of the MVAR.  Less of the voltage is going to the MW real power load. 

Same thing would happen if \$Q_C &gt; Q_L\$.  The load is capacitive and pf &lt; 1.  The capacitors supply all of the reactive power to the loads.  Now the source must supply P and \$Q_{NET} = Q_C - Q_L\$.  The source voltage is providing power to drive the MW and the extra MVAR for capacitors.  

This is not your case, but the same situation applies.  Less of the voltage would go to the MW real power load.