In balanced three-phase circuits, you'd use the distance to the load to account for the voltage drop in the phase conductor.
In single-phase circuits, with the neutral conductor size equal to the phase conductor, you'd double the distance to account for the total voltage drop (phase plus neutral).
To add to the above, the key is to use correct voltage. The voltage drop in single line of a 3-phase (balanced) system is the drop in the single phase voltage.
For example, if you have 0.2 ohms resistance in each phase condutor of a 208/120V, 3-phase system and 15A of load. The VD per line will be 15A*0.2 ohm=3V, The percentage voltage drop will be 3V/120V=.025 or 2.5%. Line to line VD will be 5.2V, still 2.5% of 208V.
Same is true, it were a delta system, even if there is no physical neutral. The calcs are still done on per phase basis, using phase voltage=line voltage/1.732.
(For purists: Above example ignores reactance for simplicity.)
You still need to take a look at the catalog. They list the single phase and 3 phase values of the impedance, which is not the same.
Once you read the correct value from the table, you multiply by twice the distance in 1 phase and simply by the distance in 3 phase.
You should be aware enough to know that as you have more applianced branched in parallel on the same circuit, you need to calculate the voltage drop chunk by chunk; other easier methods exist of course but should be a nice exercise.
Also, the impedance values are given in some catalogs for cos phi = 0.8 as well as for cos phi = 1, which is the worst case (highest resistance).
