I need to do this proof without the use of projections and cartesian equations.
Here's my attempt at proving that if $\vec k$ is non coplanar with $\vec u$ and $\vec v$ then the line supported by $\vec k$ (we call that line $(d)$) is secant to the plane $(P)$ of direction $\vec u$ and $\vec v$.
We work in the euclidian $3D$ space.
We start with the initial assumption, which is tatamount to saying that any point of the space can be expressed as $\alpha \vec k + \beta \vec u + \gamma \vec v$ where $(\alpha, \beta, \gamma)\in \mathbb R^3$.
Then we suppose $(d)$ is not secant to $(P)$. We then choose $A,B\in (d)$ and $C,D\in (P)$. We have $\overrightarrow{AB} = \alpha \vec k$ and $\overrightarrow {CD} = \beta \vec u + \gamma \vec v$. We can see that $\overrightarrow{AB}+ \overrightarrow{CD}$ correspond to any point of the space. Yet we know $\overrightarrow{AB}+ \overrightarrow{CD}$ can't attain the points that are inbetween $(d)$ and $(P)$.
Hence the result (the assumption that $(d)$ is parallel to $(P)$ leads to contradiction).
I am not sure if my proof, especially the end, is really correct. Does that even make sense to prove something that "visually obvious" ?