Show any any set of any $5$ vectors in $\mathbb {GF(3)}^3$ contains three vectors such that those $3$ vectors are linearly dependant.
I don't really know how to start, I have found an expample over $\mathbb R^3 $ that violates the above so $GF(3)$ is definitely relevant. Any hints?
In effect, you are asking about points in the projective plane over $\Bbb F_3$. In this context, this states that there are no five points with no three of them collinear. Such sets are known as arcs. Over $\Bbb F_q$ with $q$ odd it is well-known that an arc has at most $q+1$ points.
In our example, we can choose homogeneous coordinates such that four of the points are $[1,0,0]$, $[0,1,0]$, $[0,0,1]$ and $[1,1,1]$. The six lines joining these in pairs are $X=0$, $X=Y$ and those obtained from these by permuting coordinates. These six lines cover the whole projective plane so there can't be a fifth point.