With parameters: srg(v(v1)/6, 3(v3)/2, (v3)/2, 9)
Should be straightforward counting which alludes me...
Thanks!
Shay
With parameters: srg(v(v1)/6, 3(v3)/2, (v3)/2, 9) Should be straightforward counting which alludes me... Thanks! Shay 


It's been a time since I learned something about STS, but let's have a try. There are of course $\frac{v(v1)}{6}$ vertices in your srg since there are $v(v1)$ pairs when you have $v$ elements, and there are 6 pairs when you have 3 elements, so each triple is counted 6 times. Each element is contained in $v1$ pairs, so we get $3(v1)$ triples that are adjacent with a given triple, but we counted each triple twice and of course we also counted the current triple three times so the degree is $\frac{3(v1)}{2}3=\frac{3(v3)}{2}$. When we have two adjacent triples, then there are $\frac{v5}{2}$ triples that also contain the common element between these two adjacent triples and there are also 4 triples that contain one of the other elements of the first triple and one of the other elements of the second triple. This means $\lambda=\frac{v5}{2}+4=\frac{v+3}{2}$ (so it is + 3 and not 3 as you posted.) When we have two nonadjacent triples then there are 9 possible pairs such that one element belongs to the first triple and one element belongs to the second triple, so $\mu=9$. 

