One requirement for a 3d solid to be a polyhedron is that its faces are not skew polygons. That is, the vertices of each individual face lie in a plane. Triangular faces are inherently flat (because 3 points determine a plane). Quadrilateral faces may be flat (e.g. the polygon with vertices $(1,0,0)$, $(0,-1,0)$, $(-1,0,0)$, and $(0,1,0)$) or skew (e.g. replace any vertex with $(0,0,1)$). Faces with more vertices follow similarly.
My question is: Is there a useful measure of how skewed a 3d polygon is? My first guess is something involving singular value decomposition or principal component analysis and taking the smallest singular value, but a) it's not clear to me that this would be comparable between faces, that it would allow me to say "this is more bent than that", and b) doing PCA on 4 data points seems like overkill. I think taking the deviation of the vertices from the least-squares plane has similar problems (and might be equivalent to the smallest singular value).
A typical measure is the Hausdorff distance:
Any four points that form a skew quadrilateral determine a non-zero-volume convex hull of those four points, which can serve as a measure of how non-coplanar they are.
So a simple answer to "how skewed a 3d polygon is?": Calculate the volume of its convex hull. More sophisticated is to use the Husdorff distance with respect to your target/expected planar quadrilateral.