Cube standing on a corner Cube standing on a corner

This question has arisen from this post and the picture and the insights have been taken from the answers mentioned there.

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In the above picture, a cube is standing on one of its corners. Let the cone lie on a $xz \text{ plane}$ and assume $y$-axis (the red line) passes through the centre of the cone and through one of the body diagonals of the cube. Now for the cube, imagine a plane passing though the points $QS$ and parallel to the $xz \text{ plane}$.
What I thought:

I thought, after looking at a dice for some time, that the $3$ vertices $QST$ would indeed form a triangle but the $3^{rd}$ vertex will be at a different height (not in the same plane), like the triangle $QST'$ below. So when a $xz \text{ plane}$ will cut the cube, what we'll get will be some polygon which with my level of understanding and imagination wasn't possible to fathom then.

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Image on the Left: Front View; $\qquad\qquad\qquad$Image on the Right: Top View
What it turned out to be:

It turned out that all the $3$ vertices, $QST$ indeed lie on single $xz \text{ plane}$, like $QST$ above, and thus form a triangle. Similarly it's true for the $3$ vertices above these (the vertices $U,V,W$). And add to that all the $4$ $xz \text{ planes}$ upon which all these vertices lie are at intervals of equal height!

I want to know that:

*

*How can we know whether a triangle forms and not some other polygon for the vertices $Q,S,T$ on the $xz \text{ plane}$ parallel to cone's base and passing through $1$ or more of the $3$ vertices (which may be an unfamiliar figure, maybe a pentagon or some quadrilateral, as below)?


*Whether all the 3 vertices of the cube will lie on a plane parallel to $xz \text{ plane}$ and not on 2 planes parallel to $xz \text{ plane}$ 

*How to determine the height of these $xz \text{ planes}$ from one another?

*How will one calculate the centre and the radius for the obtained polygon in point 1? [Note: Point 4 has a bit to do with the linked post (in case you don't get what I'm saying)]

 A: To find lengths, one may work with another cube rotated to a more convenient orientation.
Consider a cube with side length $a$, with two neighbouring vertices at $O(0,0,0)$ and $E(a, 0, 0)$, and a body diagonal from $O$ to $D(a,a,a)$.
Question 3
In your diagram, you are looking for the height of point $Q$ (or $S$ or $T$) along the $y$-direction, relative to point $P$ in your diagram.
In my rotated cube, this would be to find the projected length of $OE$ along the $OD$ direction. One may perform this projection by dot product:

*

*The dot product of $\overrightarrow{OD}$ and $\overrightarrow{OE}$ is
$$\overrightarrow{OD}\cdot \overrightarrow{OE} = (a,a,a)\cdot (a,0,0) = a^2$$


*Dividing this by the length of the body diagonal $OD$, and calling the result $s$:
$$\begin{align*}
\overrightarrow{OD}\cdot \overrightarrow{OE} &= \left\|\overrightarrow{OD}\right\| \left\|\overrightarrow{OE}\right\| \cos\theta = \left\|\overrightarrow{OD}\right\|s\\
s &= \frac{\overrightarrow{OD}\cdot \overrightarrow{OE}} {\left\|\overrightarrow{OD}\right\|} = \frac{a^2}{\sqrt{3a^2}} = \frac{a}{\sqrt 3}
\end{align*}$$
where $\theta$ is the angle between $\overrightarrow{OD}$ and $\overrightarrow{OE}$, as in the linked Wikipedia page.
$s$ is the projected length of $OE$ along the $OD$ direction, which in your orientation is the height of $Q$ along the $y$-direction relative to $P$. Repeat the same process for other vertices of my cube, e.g. $(0,a,0)$, $(a,a,0)$, etc.
Question 4
From Q3 above we obtained $s$, the projected length of $\overrightarrow{OE}$ along the $\overrightarrow{OD}$ direction. If $r$ is the perpendicular distance from $E$ to line $OD$, then
$$\begin{align*}
OE^2 &= s^2+r^2\\
r^2 &= OE^2 - s^2\\
&= a^2- \left(\frac{a}{\sqrt3}\right)^2\\
r&= a\sqrt{\frac23}
\end{align*}$$
In your orientation, $r = a\sqrt{\frac23}$ would be the radial component of $Q$ away from the $y$-axis (or from $P$).
(If we find the exact "projected" vector with signed length $s$ and direction $\overrightarrow{OD}$, this will be the vector projection of $\overrightarrow{OE}$ onto $\overrightarrow{OD}$. Then subtract this vector from $\overrightarrow{OE}$, this will return a vector with length $r$ and perpendicular from $OD$. This is also the first steps of the Gram-Schmidt process.)
Question 1
From the same calculation as in Q3 above, points $Q, S,T$ all have the same $y$-coordinates, i.e. the same "height" from $P$. So they lie on the same plane parallel to the $xz$-plane.
The face diagonals $QS$, $ST$ and $TQ$ are all on this plane, and all have length $a\sqrt 2$. So $\triangle QST$ is an equilateral triangle on this plane.
Question 2
The cube has 3-fold rotational symmetry centred at a body diagonal.
For cross sections of a cube on a plane perpendicular to the body diagonal, the cross section shape maintains the same symmetry and may be either

*

*an equilateral triangle, or

*a convex hexagon.

*(or just a single point at the two ends $P$ and $V$)

One interactive demo of cube cross sections from Google search is https://www.geogebra.org/m/X6GYjh2U (not by me).
