I want to understand why given $d$ ($d - 1$)-spheres in $\Bbb{R}^d$ that intersect at three or more points, then their centers must lie on a ($d - 2$)-plane.
For $d = 2$ this doesn't tell you anything. For $d = 3$ I can see it intuitively: two distinct 2-spheres, call them $A$ and $B$ in $\Bbb{R}^3$ that intersect at more than 2 points must meet at a circle $S$ which lies in a plane perpendicular to the line $l$ containing the centers of $A$ and $B$. Any sphere $C$ that meets both $A$ and $B$ at three or more points must intersect with the circle $S$. Using the rotational symmetry of the circle and the sphere, and the fact that the line $l$ passes through the center of $S$, one can see easily that the center of $C$ must lie on the line $l$.
I would like to know how to generalize this argument for higher dimensions $d > 3$. I cannot visualise it because my mind's eye does not see in 4-D.
Suppose the intersection of $d$ different $(d-1)$-spheres contains three distinct points. These three points are not collinear, because they are on a $(d-1)$-sphere. Hence they span a $2$-dimensional Euclidean subspace $E\subset\Bbb{R}^d$. The intersection of this subspace with any of the $(d-1)$-spheres is a circle, the unique circle passing through these three points. The orthogonal projection of $\Bbb{R}^d$ onto $E$ then projects the center of each $(d-1)$-sphere onto the center of this circle; this shows that the centers of the $(d-1)$-spheres are all in the $(d-2)$-dimensional subspace of $\Bbb{R}^d$ that projects onto the center of this circle.