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 $p1, q1$ but not $r1$ and the product $pqr$ is an $l$ digit integer.
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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. 

