I've been trying to solve this question for quite a time but I didn't come up with anything.
Let $A = (a_{ij}) \in M_{m,n}(\mathbb{R})$ and consider the linear system $(S)$: $AX = B$, where $B \neq 0$. We suppose that for all $k \in \{1, 2, \dots, n\}$, we have $(a_{1k}, a_{2k}, \dots, a_{mk})B = 0$.
PS: the last product is clearly a vector product.
The condition $(a_{1k}, a_{2k}, \dots, a_{mk})B = 0$ is equivalent to $A^TB = 0$, as stated in the comments by Sammy Black. But then we can multiply by $A^T$ on both sides of $AX = B$ giving us $A^TAX = A^TB = 0$, implying that the column-vectors of $X$ are in $\textrm{null}(A^TA)$.
Now, since $\textrm{null}(A^TA) = \textrm{null}(A)$, see this answer, we must have that the column-vectors of $X$ are in the nullspace of $A$, thus $0 = AX = B$, which is a contradiction, thus the system must be inconsistent if $B \neq 0$.
In addition to the good first answer by @Dragonoverlord3000, here is an intuitive one.
$A X = B$ means $B$ is a linear combination of the column vectors of $A$. (Or, if you prefer, it is in the vector space generated by them).
The last equation means each column vector in $A$ is orthogonal to $B$. This is contradictory with the previous result, if $B$ is not null.
