After seeing this post Number of cyclic subgroups order $p^2$ in $\mathbb{Z}_p \times \mathbb{Z}_p \times \mathbb{Z}_{p^2}$ I have following questions.
Let $G \simeq \mathbb{Z}_p \times \mathbb{Z}_q \times \mathbb{Z}_r$ where $p,q,r$ are primes. Then how many elements of order $p$, order $q$, and order $r$?
And what if I extend this $p,q,r$ to $p^{a}, q^{b}, r^{c}$?
I will note that your subject line currently asks about subgroups, but your post asks about elements. Those are different counts. I will answer the question in the post.
If $p$, $q$, and $r$ are pairwise distinct, then $\mathbb{Z}_p\times\mathbb{Z}_q\times\mathbb{Z}_r$ is cyclic of order $pqr$, and hence has a unique subgroup of any order dividing $pqr$; so it has exactly $p-1$ elements of order $p$, $q-1$ elements of order $q$, and $r-1$ elements of order $r$.
The same is true if we take $p^a$, $q^b$, and $r^c$, for the same reason.
If two of the primes are equal and the third distinct, then we have, say, $\mathbb{Z}_p\times\mathbb{Z}_p\times\mathbb{Z}_r$. There are exactly $r-1$ elements of order $r$, and $p^2-1$ elements of order $p$. If you take $\mathbb{Z}_{p^a}\times\mathbb{Z}_{p^b}\times\mathbb{Z}_{r^c}$, it’s the same answer.
If all primes are the same, and you have $\mathbb{Z}_{p^a}\times\mathbb{Z}_{p^b}\times\mathbb{Z}_{p^c}$. Each factor has exactly $p$ elements of exponent $p$, and an element of the whole thing is of order $p$ if and only if every coordinate is of exponent $p$ and at least one is of order $p$. That gives you $p^3-1$ elements of order $p$.