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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 $(x/\log x)^3(\phi(m)-1)/\phi(m)^2$. 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.