If (u,v,w) is a family of linearly dependent vectors in vector space V and vector x is in the span of (u,v,w), then x=αu+βv+γw has infinitely-many choices for α,β, and γ.
If (u,v,w) is linearly dependent, then there exists an α, β, and γ, not all equal to zero, such that αu+βv+γw=0.
The Attempt at a Solution
My first attempt, which didn't go anywhere, was to assume that there were only a finite number of choices and see if that led to a contradiction.
For my second attempt, I started with the fact that (u,v,w) is linearly dependent. Then I multiplied both sides by an arbitrary scalar n. Then I thought I could add x to both sides and manipulate the equation somehow, but that didn't lead anywhere either.
If I was dealing with the vector space R^n, then x=αu+βv+γw would have infinitely-many solutions because you would end up with at least one free variable since at least one of the vectors is just a linear combination of the others. Can you apply that reasoning to all vector spaces?