Intersection of perpendicular chords: is this true for a sphere?

For circles:

From the circle center draw perpendicular lines to the chords. Let the intersection of the lines with the chords be $X$ and $Y$, respectively. Let $|OX|=x$, $|OY|=y$.

We have (see figure):

$$\begin{align} a&=\sqrt{R^2-x^2}+y,\\ b&=\sqrt{R^2-x^2}-y,\\ c&=\sqrt{R^2-y^2}-x,\\ d&=\sqrt{R^2-y^2}+x, \end{align}$$ so that $$ a^2+b^2+c^2+d^2=4R^2, $$ as claimed.

Try to apply the same construction for the sphere case.

Drawing from the center of the sphere the planes perpendicular to the chords and denoting the intersection points $X,Y,Z$ you will obtain: $$a^2+b^2+c^2+d^2+e^2+f^2=6R^2-2(x^2+y^2+z^2),$$ so that the claim does not hold for a sphere.