Let $m$ be a fixed integer. 
I want to count number of prime triplet $(p,q,r)$ such that $p < q < r < 2p$ with $m$ divides $p-1, q-1$ but not $r-1$ and the product $pqr$ is an $l$ digit integer. 



Counting the number of prime triplet
Under your assumptions $p,q,r$ are all about size $x= 10^{l/3}$. The congruence conditions are basically independent so you'd get about $(x/\log x)^3(\phi(m)-1)/\phi(m)^3$. There may be a constant in front to account for the inequalities among the primes and the fact that you want exactly $l$ digits. This should be OK when $l$ is large compared to $m$. If that's not the case, it might be trickier.


