How to determine if vectors span a plane?

Let $vec{u}=(1,2,-1)$, $vec{v}=(2,0,1)$ and $vec{w}=(3,2,0)$. We have $mathrm{Span}{vec{u},vec{v}}=mathrm{Span}{vec{u},vec{v},vec{w}}$. The parametric equations of plane generated by $vec{u}$ and $vec{v}$ is $$ (x(t,r),y(t,r),z(t,r))=tcdot vec{u}+rcdot vec{v} $$

If any 2 vectors in S are not multiples of each other, then those 2 vectors span a plane. But that's is easier if you have a small set of vectors to check.

If you write $S={v_1,v_2,v_3}$ and you observe $v_3=v_1+v_2$, you can see that $mathrm{span}(v_1,v_2,v_3)=mathrm{span}(v_1,v_2)$, and because $v_1neq0neq v_2$ and $v_1neq lambda v_2$ for any $lambdainmathbb{F}$, with $mathbb{F}$ the field where the scalars from your vector space come from, you can see that ${v_1,v_2}$ is linearly independent and thus span a plane.

Explicitly row reducing the the matrix with the three vectors is not necessary if you provide a good (similar to mine) argument.

Because $(1,2,-1)+(2,0,1)=(3,2,0)$ then $$text{span}({(1,2,-1),(2,0,1),(3,2,0)})=text{span}({(1,2,-1),(2,0,1)})$$ and the last two vectors are linearly independent. The generated by two linearly independent vectors are, in fact, a plane.