Let $G$ be a non-nilpotent and supersoluble finite group.
$G=Q\ltimes P$, Where $P$ is a group of order $p^{2}$($p$ is prime), all maximal subgroups of $P$ are normal in $G$, $Q=\langle a\rangle$ is a cyclic $q$-group ($q\neq p$) of order $|Q|>q$ and $C_{Q}(P)=\langle a^{q} \rangle$ Then every $2$-maximal subgroup of $G$ permutes with all $3$-maximal subgroup of $G$.
A subgroup $H$ of a group $G$ is called $2$-maximal (or second maximal )in $G$ if $H$ is a maximal subgroup of some maximal subgroup $M$ of $G$.
The point appears to be that every third maximal subgroup of $G$ is contained in $C = P Q^q = P \langle a^q \rangle$.
The reason is that if $T$ is a third maximal subgroup, its index must be divisible by $q$. So $q$ divides the index of $PT$. Now $G/P$ is isomorphic to $Q$, so by the third isomorphism theorem $T \le PT \le P \langle a^q \rangle$, because $P \langle a^q \rangle/P$ is the only subgroup of index $q$ in $G/P$
Now $C$ is a cyclic normal subgroup of $G$, so every subgroup of $C$ is normal in $G$, and thus permutes with every other subgroup.