It is known that the group you mention does not act geometrically on a CAT(0) cube complex. See for instance this answer for a possible argument, based on Lemma 16.12 in Wise's monograph *The structure of groups with a quasiconvex hierarchy*. I think you can adapt the proof of this lemma in order to prove the following statement:

**Proposition:** *Let $G$ be a group acting properly and cocompactly on a median metric space $(M,d)$ of finite rank. Assume that $G$ contains a finite-index subgroup $H \simeq \mathbb{Z}^n$, $n\geq 2$. Then $G$ acts properly and cocompactly on $(\mathbb{R}^n, \ell^1)$.*

**Sketch of proof.** As shown by Bowditch in *Some properties of median metric spaces*, there exists a CAT(0) metric $\sigma$ on $M$, and, if I understand the construction correctly, this metric satisfies the following properties: (1) any isometry of $(M,d)$ induces an isometry of $(M,\sigma)$; (2) the metrics $d$ and $\sigma$ are biLipschitz equivalent; (3) halfspaces of $M$ are $\sigma$-convex.

As a consequence, the flat torus theorem can be applied, and we find a $G$-invariant and $\sigma$-convex subspace $\Sigma \subset M$ which is $\sigma$-isometric to $\mathbb{R}^n$.

Now, we consider the structure of measured wallspace of $\Sigma$ induced by the walls of $M$. By $\sigma$-convexity of $\Sigma$ and of the walls, we must have $m$ families of parallel hyperplanes $\mathbb{R}^{n-1}$ in $\Sigma$. (Here, we ignore the collections of hyperplanes which lie in the neighborhood of a single hyperplane. As a consequence, $m \leq n$ since otherwise it would be possible to embed coarsely $\mathbb{R}^{n+1}$ into $\Sigma \simeq \mathbb{R}^n$.)

Let $F$ denote the median space associated to the previous wallspace. Then $F$ decomposes as the $\ell^1$-product of $m$ (discrete or continuous) unbounded lines. Up to replacing discrete lines with continuous lines, we may suppose that $F$ is $(\mathbb{R}^m,\ell^1)$.

Notice that, because $G$ acts properly on $\Sigma$, necessarily it also acts properly on $F$. So we must have $m \geq n$. But we already know that $m \leq n$, so $m=n$.

So far, we have proved that $G$ acts properly on $(\mathbb{R}^n,\ell^1)$. As $G$ is virtually $\mathbb{Z}^n$, we conclude that $G$ acts geometrically on $(\mathbb{R}^n,\ell^1)$.
$\square$

In your specific example, the question is now: does $\mathbb{Z}^2 \rtimes \mathbb{Z}_6$ act geometrically on $(\mathbb{R}^2,\ell^1)$? The same argument as the one followed here works. A presentation of the group is
$$T=\langle a,b,c \mid a^2=b^2=c^2=(ab)^3=(bc)^3=(ac)^3=1 \rangle.$$
Notice that $\mathrm{Isom}(\mathbb{R}^2, \ell^1)= (\mathbb{R}\rtimes \mathbb{Z}_2)^2 \rtimes \mathbb{Z}_2$ does not contain elements of order three, so, for any action $T \curvearrowright (\mathbb{R}^2, \ell^1)$ by isometries, the elements $ab$, $bc$ and $ac$ must be trivial. Consequently, such an action must factorise through the quotient $T \twoheadrightarrow \mathbb{Z}_2$ sending all the generators to $1$. A fortiori, the action cannot be geometric (and even proper).

**Remark:** The argument above does not completely answer the question as the median metric space is supposed to have finite rank. I do not know what happens if the rank is infinite, but it seems reasonable to think that the same conclusion holds.