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).
Best Answer
I think your proof is fine.
In the following, I'm going to write some comments and another proof.
Some comments : In your proof that $M,N,T$ are collinear, it might be better to explicitly say that you used the fact that given three non collinear points, only one plane goes through them. Also, it might be better to add a few explanations about why $PQSM$ is a rectangle when $PQ=MS,PQ\parallel MS$ and $PS=MQ$.
Another proof that $M,N,T$ are collinear : Using the fact that the intersection of two non-parallel planes in three-dimensional space is a line, we can say that the intersection of $\alpha$ and the plane $A'B'C'D'$ is the line $MN$. It follows that $M,N,T$ are collinear. (Also, it follows that $M,N,T'$ are collinear.)