I guess the answer should be that (a,b,c) fills the line if and only if there is an integral combination of its permutations equaling (1,1,1). More later... typing on my phone now.

I guess the answer should be that $(a,b,c)$ fills the line if and only if there is an integral combination of its permutations equaling $(1,1,1)$. We want the elementary divisors of a certain $3 \times 6$ matrix. Working out explicit conditions should be do-able. I wonder if there is a slick argument.

There is also a neat connection with semi-magic squares, I think.

When allowing longer vectors which are permutations of $(a,b,c,0,\dots,0)$, it feels like their Frobenius number is the truth, but again with some extra necessary divisibility conditions.

(Sorry if my question was elementary.)