I've been stuck thinking about this for a while.
Def. Let $G$ be a finite solvable group whose order is divisible by only three primes: $p,q,$ and $r$. Suppose that $G$ has cyclic subgroups of order $pr$, $pq$, and $qr$, but no cyclic subgroup of order $pqr$. Furthermore, (for the sake of eliminating obvious duplicate cases that are not relevant to the nature of my question,) suppose that no proper subgroup of $G$ has this property. I'm calling such $G$s sneaky groups.
The only sneaky groups that I can think of are similar, in spirit, to the following example. Let $\langle a_1 \rangle \times \langle a_2\rangle = C_2\times C_2$ act on $\langle b\rangle \times \langle c \rangle = C_3\times C_5$ so that $a_1$ commutes with $C_5$ and acts fixed point freely on the $C_3$, and $a_2$ commutes with $C_3$ and acts fixed point freely on the $C_5$. We end up with $$\left(C_3\rtimes C_2\right)\times \left(C_5\rtimes C_2\right) = \left(\langle b \rangle\rtimes \langle a_1 \rangle\right)\times \left(\langle c \rangle \rtimes \langle a_2 \rangle\right)$$ which has elements $ca_1$ of order $5\times 2$, $ba_2$ of order $3\times 2$, and $bc$ of order $3\times 5$. But, since $a_1$ doesn't commute with $b$ and $a_2$ doesn't commute with $c$, we can't have any element of order $3\times 5 \times 2$. Sneaky.
We can generalize this construction by replacing the $C_2\times C_2$ with a product of (solvable) Frobenius complements, and the $C_3\times C_5$ with a product of Frobenius kernels, with actions defined in such a way that each complement commutes with one kernel and acts fixed point freely on the others. Furthermore, we could expand one more time to using 2-Frobenius groups of type $(p,q,p)$ (see p.5 here for definition), for example by replacing the $C_3$ and one of the $C_2$s with an $S_4$.
So, my question is,
Is it true that every sneaky group is a product of Frobenius and/or 2-Frobenius groups? If there a counterexample, what is the counterexample?