Let $1\leq \ell \leq \binom{k}{2}$. It seems that there should exist some set $\mathcal{P}$ of $kd-\ell+1$ points in $\mathbb{R}^d$ which do not lie on the union of any $k$ hyperplanes $H\_1,\ldots, H\_k$, exactly $\ell$ of which are orthogonal. Indeed, this should be generic. 


If $\ell=1$, this is easily seen: put the points in general position and chosen so that the affine hulls of any two subsets of the points are not orthogonal. But this argument doesn't seem to work for 7 points in $\mathbb{R}^3$. 


Any suggestions, or reference for something similar?



7 Points in $\mathbb{R}^3$ that don't lie on the union of 3 orthogonal hyperplanes
Of course, fedja's measure theoretic argument works. Here is another way to see that $7$ general points in $\mathbb{R}^3$ don't lie on $3$ mutually orthogonal planes. For $7$ generic points we will have 


1. No $3$ on a line and no $4$ on a hyperplane.
2. For any two triples of points, the hyperplanes they span will not be orthogonal.
3. For any triple of points, the projections of the remaining $4$ points onto the hyperplane it spans do not lie on $2$ orthogonal lines.


The first condition means that we cannot put $4$ points on a hyperplane, and we may speak uniquely of the hyperplane through any $3$ points.


The second condition means that we cannot put $3$ points on one hyperplane and $3$ on another orthogonal hyperplane.


The last means that we cannot put $3$ points on one hyperplane and $2$ on each of two other hyperplanes, orthogonal to each other and to it.


