Example: Let $a$, $b$, and $c$ be a system of non-zero, non-coplanar vectors. Examine whether the vectors: $$f=5a+6b+7c$$
$$g=7a-8b+9c$$
$$h=3a+20b+5c$$ are linearly independent or dependent.
and the answer goes as follows:
Suppose $f$, $g$, and $h$ are linearly independent.
Then, $αf+βg+γh=0$ where $α=0, β=0, γ=0$
Consider $αf+βg+γh=0$ -(1), substituting $f$, $g$, and $h$ into (1) and noting that $a$, $b$, and $c$ are non-coplanar vectors, we get:
$$5α+7β+3γ=0 -eq(2)$$ $$6α-8β+20γ=0 -eq(3)$$ $$7α+9β+5γ=0 -eq(4)$$
(2) - (4) $$α+β+γ=0 -eq(5)$$
(3) & (4) substitution gives $α=-2γ$ and $β=γ$.
Hence there exists non-zero scalars such that (1) holds. This contradicts (1) hence the given vectors are linearly dependent.
My question is, how does the $α=-2γ$ and $β=γ$ show that they are non-zero?
Here's a fixed version of the answer.
By definition, $f,g,h$ are linearly independent iff the equation $\alpha f+\beta g+\gamma h=0$ has $\alpha=0,\beta=0,\gamma=0$ as its only solution. So let's assume $\alpha f+\beta g+\gamma h=0$, try to solve for $\alpha,\beta,\gamma$, and see if there are nonzero solutions. We have
$$\alpha(5a+6b+7c)+\beta(7a-8b+9c)+\gamma(3a+20b+5c)=0$$
which rearranges to
$$(5\alpha+7\beta+3\gamma)a+(6\alpha−8\beta+20\gamma)b+(7\alpha+9\beta+5\gamma)c=0.$$
Now use the fact that $a,b,c$ are linearly independent vectors (this is what is meant by "non-coplanar"). This allows us to conclude from the vector equation above that the three scalar coefficients in parentheses are all $0$. So we have the system of linear equations
$$ 5α+7β+3γ=0\\ 6α−8β+20γ=0\\ 7α+9β+5γ=0. $$
By Gaussian elimination (or just ad-hoc application of elementary row operations), we find that the above system is equivalent to the system
$$ 1α+0β+2γ=0\\ 0α+1β-1γ=0\\ 0α+0β+0γ=0 $$
or just
$$ \alpha=-2\gamma\\ \beta=\gamma. $$
Since our system with three unknowns reduces to just two equations, there will be nonzero solutions. We can choose any desired value for $\gamma$ and get corresponding values of $\alpha,\beta$ to form a solution. For example, taking $\gamma=1$, we get $\alpha=-2,\beta=1$. And indeed we can verify that
$$(-2)f+1g+1h=0$$
which shows directly that $f,g,h$ are not linearly independent.