Thank you for your answer. The $q$ entries are indeed fixed. The original question proves that $q<k(k+1)/2$ is sufficient for $M_V\cap M_X$ to have a non-trivial intersection. Your answer seems to extend the result to $q\leq k(k+1)/2$. This is indeed helpful, but what I really need is a necessary condition for $M_V\cap M_X$ to have a non-trivial intersection. Also, I am having some difficulty following your proof. You write "I claim that the column space of X has codimension at most k-1", but this assertion isn't obvious to me. Could you please explain why it is true? Thank you.

I should add that $V$ is a random subspace whose distribution is uniform over all possible $k$-dimensional subspaces of $\mathbb R^n$, while the zero constraints are independent of $V$. There are pathological choices of $V$ where $q<\frac{k(k+1)}{2}$ is not a necessary condition (e.g, we could have $M_V\subseteq M_X$). However, I think these pathological choices have probability zero.