Someone should email quid's comment (the link) to Frank's wife and delete it before he finds it out so he gets a surprise on what he thought should be a cake with Stokes' theorem.

You can do the curly braces by \lbrace and \rbrace in math mode. I learned this by right-clicking math in others' posts. You can literally see how they typed their math this way.

@Felix ...and the condition given in Edit3 can be met only when $v = k(k-1)+1$, which implies that the $S(2,k,v)$ is symmetric, be the way. So if $v > k(k-1)+1$ (i.e., the design becomes asymmetric due to $v$ being too large if you will), an $S(2,k,v)$ has a pair of parallel blocks. So, for instance, if you pick $k=3$, using a large set of $S(2,3,v)$s will give you a simple $2$-design with a pair of parallel blocks while satisfying $r = \lambda^2$. If you don't need your $2$-design to be simple, simply copying the same $S(2,k,v)$ $\frac{v-1}{k-1}$ times does the job.

@Felix For large $v$, the construction works just fine. I'll write up the details as an edit to the answer.

Ah, I didn't F5 before posting the above comment. I'm sorry.

Didn't you forget some other restrictions on the kind of (block) design you want? If you're really ok with a $2$-design of any kind as long as $r = \lambda^2$, then infinitely many of them sure exist. For example, for any finite point set $V$, you take $B = V$ as its only block. Obviously $\lambda = 1$ because every pair of elements of $V$ appears exactly once in a block of the block set (which is a singleton consisting of a block of size $\vert V \vert$). You have $r = 1$ as well because every point appears exactly once for obvious reasons.