This is an extension of this question. To rehash the main ideas, we have vectors $(0,0,0,0,0,0)$, $(a,a,a,0,0,0)$, $(b,b,b,b,b,b)$, $(c^0,c^1,c^2,c^0,c^1,c^2)$ and $(0,d,2d,0,0,0)$ in a linear algebra system.
I'm hoping that we can create a (6-dimensional) basis for the system using an additional vector: $(0,-e,0,0,0,2e)$. I obtain this vector using the other vectors, knowing there are limits to the size of each value in the vectors. I get $(c^0,c^1-d,c^2-2d,c^0,c^1,c^2)$ by adding two of the original vectors. Then I make $c$ very large. The idea is that I then set $2d=c^2$, and that cancels out one of the values in the new vector, so that I'm left with $(c^0,c^1-d,0,c^0,c^1,c^2)$. Since $c$ and $d$ are very large, it follows from a little bit of thought that this new vector is approximately $(0,0-d,0,0,0,c^2)$. Again, from $2d=c^2$, we have this roughly equal to $(0,-d,0,0,0,2d)$. Does this help to form a 6-dimensional system? In other words, can I determine all 6 indices from these vectors?