That optical isolator tells me that the designer intended the left half of this circuit to operate completely isolated from the right. That's a good design choice, since the left side is connected to mains.
I can't read the IC designators for the opamps, but I assume those two opamps are from different physical packages. All opamps in the same package share the same power supply, they are all physically connected to the same supply rails inside the package. If those two units in the schematic come from the same package, that entirely defeats isolation.
I am fairly certain that the left opamp is housed in one IC, and the right opamp is in another completely different IC, and that each IC is powered from its own independent, isolated supply.
The left opamp is probably powered from the supply that provides VCC2 and GNDA, (the one that you also show us), since VCC2 is used on the input side of the opto-isolator.
The right opamp is probably powered from a completely separate and isolated source of +12V, as suggested by its use on the output side of the optical isolator, a supply which we don't see here.
The left opamp is a differential amplifier, and responds to the difference in potential between mains live and neutral, there's no need to connect mains neutral or live anywhere other than shown.
GND for the +12V side of the circuit should not be connected to GNDA from the VCC2 supply side. The whole idea of the opto-isolator is to eliminate the need for such connections.
VCC2 is derived from a bridge-rectified AC source. The use of a diode bridge means that connecting either side of that source (neutral or live) to GNDA or any other point in the DC side of the circuit will make fire.