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I think there are examples when the Borel subgroup is not maximal solvable in a parabolic. One such is when $q = p^{a}$ and $p \leq 3,$ so for example when $G = {\rm GL}(3,2)$ both the maximal parabolics are themselves solvable (isomorphic to $S_{4}$). Similarly when $p= 3$, the group ${\rm GL}(3,3)$ has a solvable maximal parabolic $P$ with unipotent radical $U$ such that $P/U \cong {\rm GL}(2,3).$ Such behaviour perpetuates itself in higher ranks to give solvable parabolic subgroups strictly containing the Borel in characteristic $2$ or $3$.