If you think about matrices as coefficients of systems of linear equations, $A$ are the coefficients of a system of $3$ linear equations in two unknowns, while $B$ corresponds to a system of 2 linear equations in 3 unknowns.
That means that the system corresponding to $B$ is under-determined, and as such, the homogeneous system $B\mathbf{x}=\mathbf{0}$ always has nontrivial solutions. So you can find $\mathbf{a}\neq\mathbf{0}$ such that $B\mathbf{a}=\mathbf{0}$. That means that $(AB)\mathbf{a}=A(B\mathbf{a}) = A\mathbf{0} = \mathbf{0},$ so the system of 3 linear equations with 3 unknowns given by $AB$ has nontrivial solutions.
But if the coefficient matrix $M$ of a system with the same number of equations as unknowns is invertible then the only solution to $M\mathbf{x}=\mathbf{0}$ is the trivial solution: for multiplying by $M^{-1}$ we get $\mathbf{x}=M^{-1}M\mathbf{x} = M^{-1}\mathbf{0}=\mathbf{0}$.
Since the system of 3 equations in 3 unknowns $(AB)\mathbf{x}=\mathbf{0}$ has a nontrivial solution, then $AB$ cannot be invertible.
(If you know the rank-nullity theorem, and that a square matrix is invertible if and only if its nullity is $0$, you can do the argument above using nullities, noting that $\mathbf{N}(B)\subseteq \mathbf{N}(A)B$, so $\mathrm{nullity}(AB)\geq \mathrm{nullity}(B)$, and that $3 = \mathrm{rank}(B)+\mathrm{nullity}(B)$ and $\mathrm{rank}(B)\leq 2$.)