When does the set of integer linear combinations of 3 vectors in the plane form a dense subset? Is there an easy to check if and only if condition for when the set of integer combinations of 3 vectors in the plane will form a dense subset of the plane? It seems like having no vector being a multiple of another, and having all 3 vectors being linearly independent over the field ${\mathbb Q}$, and not having any coordinate where all 3 vectors are rational multiples of one another is a necessary condition, but I'm not sure if it's sufficient.
 A: As I said, we can demand the first two vectors to be the standard coordinate vectors, so they give the standard integral lattice. Then, on page 28 of Diophantine Approximations by Ivan Niven we find 
Kronecker's Theorem: Given irrationals $\alpha, \beta$ such that  $1,\alpha, \beta$ are linearly independent over the rationals. Then
 the points $$ ( m \alpha - \lfloor    m \alpha \rfloor \; , \; \;  m
 \beta - \lfloor    m \beta \rfloor )  $$ for $m = 1,2,3,4, \ldots$ are
 dense in the unit square.
Proof takes pages 28-31.
As a result, the integer coefficient sums of the three vectors are dense in the plane. 
NEXT DAY: Kronecker's hypothesis is stronger than just $\alpha, \beta, \beta/\alpha$ irrational. For example, with $\alpha = \pi, \beta = \pi + 1,$ which does give a dense winding line but fails Kronecker's hypothesis, any $(m \alpha, m \beta) = (m \pi, m \pi + m )$ is equivalent, mod the lattice, to $(m \pi, m \pi  ),$ which is, in turn equivalent to some point $(t,t)$ in the fundamental region $0 \leq x < 1, 0 \leq y < 1.$ Not only are these points not dense in the square, they all lie on a single line segment. For $\alpha = p \pi, \beta = q \pi + r$ with nonzero integers $p,q,r,$ all of Kronecker's points lie on a finite number of parallel line segments, equally spaced.
