Prove that any vector space given by the solution set of $ax + by + cz = 0$ in $\mathbb{R}^3$ for $(a, b, c)\not = (0, 0, 0)$ has dimension $2$. Hint: use the rank-nullity theorem.
The way i would do the question is by substituting $x$ and $y$ to be $s$ and $t$, then, making it
$$cz=-as-ut,\qquad z=-(as+ut)/c $$ And the basis would be $\{(1,0,-a/c),(0,1,-b/c)\}$, thus the dimension would be $2$
However, i am not sure what to do if i have to use the rank-nullity theorem. Many thanks.
You don't need to rename those variables.
Your solution has only one problem, namely it assumes $c\ne 0$. But we can overcome this, saying that at least one of $a,b,c$ is nonzero by hypothesis, and then we can express $x,y$ or $z$ from the others in a similar manner as you did, arriving to a 2d subspace of $\Bbb R^3$.
With rank-nullity theorem, consider the linear map $\Bbb R^3\to\Bbb R$ $$(x,y,z)^T\,\mapsto\,ax+by+cz\,.$$ Its rank will be clearly one.