3-Col using each colour exactly $|V|/3$ times
No, the problem is still NP-complete.  Suppose you want to know if $G$&nbsp;is 3-colourable (with no limits on how many vertices can be each colour). Just make a new graph $G^+$ by adding $2|V(G)|$ new vertices with no edges between them, so $|V(G^+)|=3|V(G)|$. If $G$&nbsp;has a 3-colouring then you can get a "balanced" 3-colouring of&nbsp;$G^+$ by setting the colours of the new vertices appropriately; if $G$&nbsp;has no 3-colouring, then $G^+$ also has no 3-colouring, so it has no balanced 3-colouring, either.