TITLE: Coin problem with permutations
QUESTION [13 upvotes]: Let $a,b,c$ be positive integers with gcd$(a,b,c)=1$, and let $\mathbb{N}$ denote the set of nonnegative integers.
It is well known that $\mathbb{N} \setminus (a \mathbb{N}+b \mathbb{N} + c \mathbb{N})$ is a finite set, but it seems difficult to determine a formula for its maximum (the Frobenius number).  This is the famous 'coin problem'. 
I am asking about a related problem.  Consider now the set of vectors $$(1,1,1) \mathbb{N} \setminus \sum (a,b,c) \mathbb{N},$$
where the sum ranges over all six permutations of $(a,b,c)$.  Is it necessarily the case that, for relatively prime $a,b,c$, there is a maximum constant integral vector in this set?  If so, what can we say about the maximum?
I suspet this may be connected (via a clever argument) with the ordinary postage stamp problem, but a quick 10-minute effort on my part failed.
edit: (here come the hypotheses!)  $a+b+c$ must be odd
If this answer turns out to be negative, or the question turns out to be hard, let's make it easier by increasing the number of coordinates: $(1,\dots,1) \mathbb{N} \setminus \sum \mathbf{v} \mathbb{N}$, where the sum is over all $\mathbf{v}$ with one entry of each of $a$, $b$, $c$, and the rest $0$.  Does this modification of the problem start to approach the ordinary coin problem for $a,b,c$?

REPLY [2 votes]: 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.)