Your answer contains quite a few leaps of logic I'm afraid. Let's go over your arguments in detail.
$\rightarrow$ If $b=0$ then the system has at least one solution
which is a trivial one
This is correct.
or, moreover, infinite number of solutions (including the trivial one).
While this is true for infinite fields, it is not true for a finite field.
And what's worse is that you are just asserting this without any proof. You'll have to give an argument for why this is the case.
Clearly, in both cases the solutions set is a linear subspace of $\mathbb R^n$
True (and obvious) if $0$ is the only solution. But there are plenty of infinite subsets of $\mathbb R^n$ that are not subspaces.
(it includes the null space and closed under additivity and scalar multiplication).
This is the part you actually need to show.
Alternatively,
If $b\neq 0$ then has no solution or has a unique solution, both way is a vector of constant terms which don't satisfies the condition of a linear subspace.
In general there might be a lot of solutions of $Ax=b$ even if $b\neq 0$. For example, if $A=\left(\begin{matrix} 1 & 0\end{matrix}\right)$ and $b=1$ then all vectors whose first component is $1$ are solutions. In general the solutions of such an equation form a so called affine subspace.
So how do we prove it? Let us call the set of solutions $S$. We want to show that $S$ is a linear subspace if and only if $b=0$.
$\Rightarrow$: Let us assume that $S$ is a linear subspace. Then $0\in S$ which means that $A0=b$. Therefore $0=A0=b$.
$\Leftarrow$: Suppose $b=0$ we want to show that $S$ is a linear subspace. So we need to show that $S$ is closed under addition and scalar multiplication. For addition assume that $x,y\in S$. This means that $Ax=0$ and $Ay=0$. Therefore $A(x+y) = Ax +Ay=0+0=0$, which means that $x+y\in S$.
I'll leave it up to you to show that $S$ is also closed under scalar multiplication.
This idea has a lot of relations to differential equations.
So, let $AX=0$ and $AY=B$ for some particular $X,Y$.
Note that $cX$ is a general solution to $AX=0$ since any multiple of $X$ should also be in the nullspace.
$$A(cX+Y)=A(cX)+AY=cAX+AY=c\quad\!\!\!\!\!\!\cdot 0+AY=B$$
So it follows that $cX+Y$ is the general solution to $M$.
Best Answer
An homogenous system of linear equations alwwyas has a solution: the null solution.
In your case, you have three equations in three unknowns. Then your vectors span $\mathbb{R}^3$ if and only if the system has one and only one solution.