So R114 sets up a biasing operational point? or how does it balance the input current?
Why would they want a 2.15MΩ resistor in the feedback loop? this makes the gain stable?
I don't see Meg ohms resistors in the feedback loop often in gain stages, I'm not sure why it's used and for what reason
1.) Although the opamp input resistances are neglected during calculations, there is a small DC current into both input terminals. And it is the task of R114 to develop a small DC voltage (caused by the small DC input current into the non-inv. terminal). This DC voltage should be approximately equal the DC voltage developped at the inv. terminal). Thus, the remaining DC differential voltage is rather small (it is a kind of input offset). As a consequernce, the influence on the desired operating point is rather small. This method is called "input current compensation".
2.) It is not possible to isolate the Meg-ohm resistor from all other elements and to ask "why so large"? This resistor is part of the overall feedback network (which - in fact - it is a bridged-T topology). All elements of this RC-feedback network determine the filter response. If you want to verify all the component values, you have no other chance than to analyse the filters transfer function. If you are interested, I can write down this function. From this function one can derive gain, pole frequency and quality factor of the filter.