As I understand, there is a $3$-dimensional analogue of Segre's theorem stating that the maximum size of a set in ${\bf F}_q^3$ ($q$ odd) with no three points collinear is $q^2+1$. I am trying to bound the size of a set in ${\bf F}_q^3$ such that there do not exist three points $$(x_1, y_1, z_1),(x_2, y_2, z_2),(x_3, y_3, z_3)$$ in the set such that $$(x_1, z_1, y_1),(x_2, y_2, z_2),(x_3, y_3, z_3)$$ are collinear (note that one of the points has been "twisted" by interchanging the second and third coordinate). My thought (and hope) is that such a set must be quite a bit smaller, because for any set of three points, there are sort of "three chances" for the points to break the rules (three ways of choosing which point to twist). But the question looks so similar to the first one that I am having doubts... perhaps there is actually a cardinality-preserving bijection between sets of the first sort and sets of the second sort?

In fact, for my application I would like to disallow subsets $\{(x_1, y_1, z_1),(x_2, y_2, z_2),(x_3, y_3, z_3)\}$ of points with $$(x_1, z_1, y_1),(x_2, y_2, z_2),(x_3, y_3, z_3)$$ collinear (as above) unless $x_1 = x_2 = x_3$, the $y_i$ are all distinct, and the $z_i$ are all distinct. This seems slightly more complicated, since we're allowing some points back into the set, so perhaps it's best to get a sense of the simpler problem first.

EDIT: I just realised that it might be easier to say something about a polynomial that vanishes on such a set. Must it have total degree $\geq d$ or perhaps the degree of some variable $x_i$, $i=1,2,3$ must be greater than $d$? I would be even more interested in any result of this kind.