Given 3 distinct primes {$p,q,r$}, then $|G|=pqr \implies G$ not simple

Let $G$ be group of order $pqr$, where $p, q$ and $r$ are primes and $p < q < r$.

You can show that there is a normal Sylow subgroup as follows. Let $n_p, n_q$ and $n_r$ be the number of $p$-Sylow, $q$-Sylow and $r$-Sylow subgroups, respectively. Assuming $n_p, n_q, n_r > 1$ implies by Sylow's theorem that $n_p \geq q$, $n_q \geq r$ and $n_r \geq pq$. Counting elements in Sylow subgroups then gives you more than $pqr$ distinct elements, which is a contradiction. Hence there exists a unique Sylow subgroup for some prime dividing $|G|$, which has to be normal.

Actually more is true: using the fact that there is a normal Sylow subgroup, you can show that the $r$-Sylow subgroup has to be normal. This result generalizes to groups of squarefree order. In a group of squarefree order the Sylow subgroup corresponding to the largest prime divisor of $|G|$ is normal. This fact can be shown with Burnside's Transfer Theorem. A more elementary proof, but not so "well-known" these days can be given using Frobenius theorem, which states that number of solutions to $x^n = 1$ in $G$ is a multiple of $(n, |G|)$.