Is the domination number of a combinatorial design determined by the design parameters? Let $D$ be a $(v,k,\lambda)$-design. By the domination number of $D$ I mean the domination number $\gamma(L(D))$ of the bipartite incidence graph of $D$.

Is $\gamma(L(D))$ determined only by $v,k$, and $\lambda$,
  irrespective of the actual structure of $D$?

I can prove this for finite projective planes and have empirically verified this property for $(8,4,3)$ and $(10,4,2)$-designs (of which there are four and three non-isomorphic ones, respectively).
P.S.
Has the domination number of such graphs been studied at all? I could only find one paper by Laskar et al. which considered the line graphs of the incidence graphs.
 A: Gordon has done a proper search of $(15,3,1)$-designs.  I guess my incorrect reasoning does lead to a computer-free proof for (15,3,13)-designs.  This is kind of cheating though, because there are repeated blocks if one takes 13 copies of a STS. The idea may work for smaller $\lambda$; see my own comment below.
Following up on Gordon's comment, consider the projective Steiner triple system $PG_3(2)$ on 15 points.
Concretely, the points can be presented as the nonzero binary 4-tuples; blocks are the triples of vectors with zero sum.
Now, the points are dominated by a parallel class of 5 blocks:
$$
\begin{array}{ccc}
0001 &0010 &0011\\
0100 &1000 &1100\\
0101 &1010 &1111\\
0110 &1101 &1011\\
0111 &1001 &1110\\
\end{array}
$$
The blocks are dominated by a maximal subspace of 7 points (e.g. those quadruples with a leading zero).
What's more, one block of the above parallel class can be taken inside the subspace.  So I think we get domination number $\le 11=(5-1)+7$.  It is going to be hard (see below) to do this well in general.  (Here I was very wrong!)
It is easy to see that, in a Steiner triple system of order $v$, covering all points with $v/3$ blocks is best possible and can occur if and only if there is a parallel class.  Likewise, touching all blocks with $(v-1)/2$ points is best possible and occurs if and only if they form a flat.  (A quick counting argument is needed.)
This is the key issue it turns out:  I acknowledge it is not correct for me to separately consider points and blocks in your bipartite graph.  That is, I have not checked carefully whether not having to dominate the chosen points by blocks, and vice-versa, fails to help enough for one of these "bad" systems. 
