Pumping lemma for $\{110^n10^{n-m}1^m: 1\le m<n\}$
You're basically there.  Taking $w=110^{2p}10^{2p−q}1^q$, there are actually three cases (you missed the first one).


$x=\varepsilon$, $y=110^{p-2}$, $z=0^{p+2}10^{2p-q}1^q$.
$x=1$, $y=10^{p-1}$, $z=0^{p+1}10^{2p-q}1^q$.
$x=110^r$ for some $r\in\{0, \dots, p-3\}$, $y=0^{p-r-2}$, $z=0^{p+r+2}10^{2p-q}1^q$.


In all three cases, taking $i=0$ means that the string $xy^iz$ is not in&nbsp;$L$.  I'd suggest trying $i=0$ and $i=2$ as your first two guesses in pumping lemma proofs&nbsp;&ndash; one of them usually works.

Notice that nothing above actually depends on the choice of&nbsp;$q$, so you can simplify things very slightly by just taking $q=1$ (or any other fixed, legal value).