Note that if $L = [p-1,q-1,r-1]\mid pqr-1$ with $p,q,r$ are distinct. then the conclusion follows trivially, so we establish the other direction. Our argument is through contrapositive. Assume either at least two of $p,q,r$ are equal, say $p=q$. Taking $n\equiv p\pmod{p^2}$, we immediately obtain a contradiction as $p^3\mid n^{p^2}$ whereas $p\mid \mid n$. Lastly, taking $n\equiv g_i\pmod{i}$ for $i\in\{p,q,r\}$, where $g_i$ is a primitive root modulo $i$, we find that $p-1\mid pqr-1$, $q-1\mid pqr-1$ and $r-1\mid pqr-1$, so the conclusion. Note. Above I omitted some minor details.