Show that $A, X, Y, Z$ are on the same plane Let $ABCD$ a rectangle and $MA \perp (ABC)$, $AX\perp MB$, $AY\perp MC$, $AZ\perp MD$. I proved that $MB\perp (XYZ)$. I have to show that $A, X, Y, Z$ are on the same plane. I attached a draw.
 A: Let's notice first that
$$MX \cdot MB = MY\cdot MC = MZ \cdot MD = MA \cdot MA$$
so $X$, $Y$, $Z$, $A$ are the transforms of $B$, $C$, $D$, $A$ under an inversion.  Now, an inversion does not take in general points on a plane into points on a plane. But what it does is, it transforms spheres into spheres or planes, and a plane into a sphere. That means that it will transform circles ( in space) into circles. Why? Because a circles is the intersection of two spheres ( or a plane and a sphere).  Therefore, since $ABCD$ lie on a circle, so will $AXYZ$.
$\bf{Added:}$ This is related to the stereographic projection. We have the projection of the plane $ABCD$ onto the sphere of diameter $MA$, and this projection takes circles in the plane to circles on the sphere.
$\bf{Added:}$ Here is another solution using vectors in 3D. Consider as origin the point $A$. Denote the vectors $\overrightarrow{AB}$, $\overrightarrow{AD}$, $\overrightarrow{AM}$ by $e$, $f$, $g$. Note that $\langle e,f\rangle = \langle e,g\rangle = \langle f,g\rangle = 0$.
We get
$$\overrightarrow{AX} = \frac{ \langle g,g\rangle e + \langle e,e \rangle g}{ \langle g,g\rangle + \langle e,e \rangle}\\
\overrightarrow{AZ}=\frac{ \langle g,g\rangle f + \langle f,f \rangle g}{ \langle g,g\rangle + \langle f,f \rangle}\\
\overrightarrow{AY}=\frac{ \langle g,g\rangle(e+ f) + \langle e+f,e+f \rangle g}{ \langle g,g\rangle + \langle e+f,e+f \rangle}
$$
Now, to show that the points $A X Z Y$ are in the same plane, it is enough to show that the vectors $\overrightarrow{AX}$,$\overrightarrow{AY}$, $\overrightarrow{AZ}$ are linearly dependent. But we have
$$ (\langle g,g\rangle e + \langle e,e \rangle g) + (\langle g,g\rangle f + \langle f,f \rangle g) = \langle g,g\rangle(e+ f) + \langle e+f,e+f \rangle g$$ and we are done.
Using the above formulae, we can check that
$$\overrightarrow{AX},\overrightarrow{AZ} ( \textrm{ and } \overrightarrow{AY})\perp \overrightarrow{MC}= e+f - g$$ so the plane $A XYZ$ is the perpendicular plane through $A$ to the line $MC$.  Perhaps this would give an idea for a synthetic solution for the problem.
