Problem: Let $T\subseteq \{0,\ldots, k-1\}$ and let $\mathcal F\subseteq [n]^{(k)}$ (subsets of size $k$) such that $|A\cap B|\in T$ for $A,B\in \mathcal F$, $A\ne B$. Show that if $n\ge k\binom{3k}{k}$ then $|\mathcal F|\le \prod_{t\in T}\frac{n-t}{k-t}$.
Notes: The classic Erdős-Ko-Rado Theorem has $T$ equalling $\{t,\ldots, k-1\}$ for some $t$, and we get the bound $\binom{n-t}{k-t}$. We can proceed similarly to the proof of EKR (see p. 25 here for instance: https://dl.dropboxusercontent.com/u/27883775/math%20notes/part_iii_combo.pdf) When we try to apply the induction hypothesis, we are in the following situation. Induct on $k$ and then on $|T|$. Write $T=\{t\}\cup S$ where $t=\min T$. If there are no $A,B$ with $|A\cap B|=t$, we are done by the induction hypothesis. If there are, and not all sets contain $A\cap B$, proceed as in EKR, applying the inequality $n\ge k\binom{3k}{k}$. If all sets contain $A\cap B$, consider $\mathcal F\backslash (A\cap B)$, and apply the induction hypothesis to get there are at most $$ \prod_{s\in S\cup \{0\}}\frac{(n-t)-(s-t)}{(k-t)-(s-t)}$$ sets. The problem with this approach is that it fails when $t=0$, and we need a separate argument there.
I think you can use the following argument.
If $0 \in T$ (i.e. $t=0$)
For each $x \in [n]$, the family $$G_x = \{ f \in \mathcal F : x \in f \}$$
also have the property that, $$|G_x| \le \prod_{t\in T\setminus \{0 \}}\frac{n-t}{k-t}$$
by the induction hypothesis (since their intersection is non-empty).
Also note that (by double counting),
$$\sum \limits_{x \in [n]}^{} |G_x| = k |\mathcal F|$$
So,
$$|\mathcal F| =\frac{1}{k} \sum \limits_{x \in [n]}^{} |G_x| \le \frac{1}{k} n \prod_{t\in T\setminus \{0 \}}\frac{n-t}{k-t} = \prod_{t\in T}\frac{n-t}{k-t}$$
So,
$$|\mathcal F| \le \prod_{t\in T}\frac{n-t}{k-t}$$