It's really a combinatorics question at heart.
For the first set of coefficients just pick (1,1,1). Whatever the result is, the number of possible triples (x,y,z) is finite -- they're constrained to the first octant because they're natural numbers, and only a finite number of lattice points on the plane x+y+z=N are in the first octant. Denote the set of possibilities for (x,y,z) by (S). Then all we need to do is find numbers (a,b,c) so that (a(x_1-x_2)+b(y_1-y_2)+c(z_1-z_2)) is never equal to zero for distinct points ((x_1,y_1,z_1),(x_2,y_2,z_2)\in S). in other words, we need a vector (a,b,c) that's not orthogonal to any of the points ((x_1-x_2,y_1-y_2,z_1-z_2)).
And this is pretty easy to find. Here's an algo, though definitely not the most efficient one. Denote the set of points ((x_1-x_2,y_1-y_2,z_1-z_2)) by (T) and let (k) be the number of elements in (T). Then simply search the space ([1,p]\times [1,p]\times [1,p]) where (p=\lceil \sqrt[3]{k}\rceil). Among those vectors there must exist one that's not orthogonal to any of the points in (T) because no two of them can be orthogonal to the same element in (T) (the angle between any two of them is strictly (<90^\circ)) and there are more than (k) of them.
Edit: Oops, that's wrong. This fixes it... Let (T) be as in the crossed out paragraph and let (a=1). Then let (m) be the largest absolute value among all (ax), where (x) is an (x)-coordinate in (T). Let (b=m+1). Then for any ((x,y,z)\in T), (ax+by\neq 0) whenever ((x,y)\neq 0). Now let (p) be the largest absolute value among all (ax+by), where ((x,y,z)\in T). Let (c=p+1), and consider an arbitrary ((x,y,z)\in T).
1. If ((x,y)=0), then clearly (z\neq 0), so (ax+by+cz\neq 0).
2. If ((x,y)\neq 0), then (ax+by\neq 0) by construction, so even if (z=0), we still have (ax+by+cz\neq 0), and if (z\neq 0), then by construction (cz) cannot equal (ax+by) in absolute value, and thus (ax+by+cz\neq 0)