How to find the supremum over all the “good” (interior) polytopes for a given set of 3D points?

Question

Let $S \subset \mathbf{R}^3$ be a set of points in 3D and let $O=(x_0,y_0,z_0)$ be the origin/point of reference.

We consider a convex polytope $P$ good / interior if:

1. $P$ is wholly contained within the interior of the convex hull of $S$: $P \subsetneq \text{convexhull}(S)$.

2. $O$ is contained within the interior of $P$, namely: $O \in P$.

3. none of the points in $S$ are in $P$ namely $S \cap P = \emptyset$.

Out of all good convex polytopes, we want to find either the maximum (supremum) of their volumes, i.e. $\sup \text{Vol}(P)$ where $P$ ranges over good polytopes.

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Here is a 2D modest illustration (the green polytope might not be really the optimal):

The black dots are the set $S$, the origin $O$ is in the middle. The red lines go through the points in $S$ which define the boundary of $S$ (not a good polytope) and the green lines go through a good polytope.

Is there an efficient algorithm to compute this best convex and non-convex polytopes for a given set $S$ of points and origin $O$?

I would like to formulate this problem as an optimization problem or to come up with some algorithm using triangulation, $\alpha$-shapes or $\beta$-skeleton geometric transformation + convex hull, linear programming, maybe working in the dual space? etc.

One idea which I have is "pumping down" the convex hull of Delaunay Triangulation:

DT $\leftarrow$ delaunayTriangulation($S \cup O$)

CH $\leftarrow$ convexHull(DT)

while Not empty(DT) and $O \notin CH$:

$\quad$ i. previousDT $\leftarrow$ DT

$\quad$ ii. update DT $\leftarrow$ DT $\setminus$ CH

return previousDT

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EDIT: I think what I am looking for in a one-liner of math symbols is:

$$\text{argmax}_{P} \text\{\text{Vol}(P) \, | \, O \in P \setminus \partial P \, ; \, S\cap P = \emptyset \, ; \, P \subsetneq \text{convexhull}(S) \} $$

Where $\partial P$ is the boundary of a polytope $P$.