Let $k$ be a field of characteristic zero and let $A$ be a finitely generated $k$-algebra. Let $B=A[x_1,\ldots,x_n]$ be the polynomial ring over $A$ and let $I \subseteq B$ be an ideal such that $B/I$ is flat over $A$. Let $f_1, \ldots, f_r \in k[x_1,\ldots,x_n]$ and let $J \subseteq B$ be the ideal generated by $I$ and the $f_j$. Is it true that $B/J$ is then also flat over $A$?
edit: Mohan pointed out that the answer is trivially "no". Does the following assumption change anything: $A\cap J=(0)$ and the $f_j$ are linear?
The answer is no. Why did you expect that? For example, consider, say $A$ is a domain and $B=A[x]$, $I=(x-a)B$ for some $0\neq a\in A$. Then the natural map $A\to B/I$ is an isomorphism and in particular flat. If $J=I+xB$, clearly $A\to B/J=A/aA$ is not flat.