Strong force potential dependence on color charge?

I have found this formula for the potential of the strong force between quarks:

V(r) = −(4/3)α(r)ℏc / r + kr

However, nothing is mentioned about color. The above is always attractive. It seems to me, for sets of three color charges to neutralize, the effect between some pair must be repulsive.

Let's assume we're in the range where the +kr term dominates. So V(r) ~= kr. The force on a test quark, which let's say has green color, from a hadron of balanced colors, should be zero. So my guess is that k is different for different pairs of colors, and negative for some pair.

Vtot(r) = k[rg]r + k[gg]r + k[bg]r

= (k[rg] + k[gg] + k[bg])r

k[rg] + k[gg] + k[bg] = 0 so that force is zero.

Possible solutions:

k[rg] > 0

k[gg] = -2k[rg] = -2k[bg]

Unlike colors attract, and like colors repel twice as strongly.

k[rg] < 0

k[gg] = -2k[rg] = -2k[bg]

Unlike colors repel, and like colors attract twice as strongly. But that's inconsistent with a hadron holding together, since each quark is affected only by unlike color quarks.

k[rg] =/= k[bg]

For example

k[rg] > 0

k[gg] = -k[rg]

k[bg] = 0

That is red and green quarks attract, like quarks repel, and blue quarks have no effect on green quarks. It seems wrong and also it turns out it's impossible to get a neutral effect on all colors of test quarks unless the red-red and/or blue-blue coupling constants are different from the green-green, which seems even more wrong. The same is true even if k[bg] is not zero but is any value different from k[rg].

I know that color is not a permanent property of quarks and that they change color as they interact, but I'm not really sure what to make of that. I also don't understand non-abelian gauge groups or whatever, so if that is the only way to explain the problem, never mind.

Is there any sense to my idea that k[gg] = -2k[rg] = -2k[bg]?