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.)