Several thoughts:
1) The MWBC approach cuts voltage drop but costs redundancy. A single 3 pole circuit now feeds 3 bays of lights, rather than 3 separate single pole circuits each feeding 1 bay. As a design decision if I would use MWBCs I would 'distribute' each MWBC so that loss of one circuit does not kill light to adjacent bays. I might even arrange things so that adjacent lights were not on the same circuit.
2) On this site I learned a clever trick for dealing with voltage drop on long circuits with distributed loads.
Rather than selecting wire size and calculating voltage drop, the approach is to select a target voltage drop and then calculate the 'circular mils' needed to feed each load with that target voltage drop. Then for each stretch of circuit you add up the circular mils required by all the loads fed through that bit of circuit to get the required wire size for all the loads.
You use the voltage drop equation from this page:
https://www.mikeholt.com/technnical-voltage-drop-calculations-part-one.php
But swap the CM (circular mil) term and the VD (voltage drop) term
CM = 2 * K * I * D / VD
K is the resistivity constant, here 12.9 circular mil * ohm / feet different values might be needed for lower temperature operation
The above formula is the single phase formula where D is the one way distance, thus the factor of 2
Here is an example:
9 lights in a bay, fed from the center.
Each light draws 1A at 265V
Target voltage drop of 12V
closest light is 300 feet from source, then 40 feet to each additional light
single pole circuit
light 1: 460feet CM= 2 * 12.9 * 1 * 460 / 12 = 989 (This means that you only need a copper cross section of 989 circular mils to feed this single lamp with the target voltage drop)
light 2: 420feet CM= 903
light 3: 380feet CM= 817
light 4: 340feet CM= 731
light 5: 300feet CM= 645
light 6: 340feet CM= 731
light 7: 380feet CM= 817
light 8: 420feet CM= 903
light 9: 460feet CM= 989
All of the lights are fed by the run from the panel to light 5. So that run needs wire size 645 + 2*731 + 2*817 +2*903 + 2*989 = 7525 circular mils
That wire size is bigger than #12 but smaller than #11. You can't buy #11 so you need #10.
The run from light 5 to light 4 carries the current for lights 1-4. So that wire needs to be 989 + 903 + 817 +731 = 3440 circular mils
That wire size is smaller than #14
Based on the above, I would consider using 15A circuits so that I could use 14ga wire. Increase the wire size only from the panel to the center light of a bay, then use 14ga for the rest.
-Jon