Not an answer, but won't fit in a comment:
This question requires us to be quite precise about what it means for some number of points to "determine a cube". It appears that @amd in his answer assumes a definition something like:
Let $C$ be the "standard cube", that is, $\bigl([0,1]^3\setminus (0,1)^3\bigr) \subseteq \mathbb R^2$.
Let $A$ be some subset of $C$. We define that $A$ "determines the cube" if for any two similarity transformations $f$ and $g$ it holds that if $f(A)=g(A)$ then $f(C)=g(C)$.
Under this interpretation, amd's answer of three points ("he diagonal across one face and a perpendicular edge at either end of the diagonal") shows that $A$ can be as small as $3$ points, and it's fairly clear that no smaller $A$ can work.
[For the parallelepiped case amd alluded to, we would presumably speak about non-degenereate affine transformations instead of similarity transformations].
However, a different interpretation of the question would be
Say that a set $B\in\mathbb R^3$ "determines a cube" if
There is a similarity transformation $f$ such that $B\subseteq f(C)$, and
For any two similarity transformations $f$ and $g$ with $B\subseteq f(C)$ and $B\subseteq g(C)$, it holds that $f(C)=g(C)$.
What is the smallest $A$ that determines a cube.
In this interpretation, three points are not enough -- for any three points there will be an infinity of large cubes that have those three points lying in the middle of one side.
This second question seems more natural to me. It can be rephrased equivalently as
What is the smallest set $A\subseteq C$ with the property that if $f$ is a similarity transformation with $A\subseteq f(C)$ then $f(C)=C$?
Unfortunately I have no idea where to start answering this. Intuitively I think that at least the eight corners of $C$ ought to work as $A$ -- that is, $A=\{0,1\}^3$. If true that would give an upper bound, but it's not clear to me how to prove it. We can't just declare it to be obvious because the analogous property in 2D is not true: The four corners of a square do not determine a unique square:
Actually, you already did most of the job.
Multiply the equation of the first plane by $\sin\varphi$, the second one by $\cos\varphi$, and add both together. That would be your single equation with a parameter ($\varphi$).
$$y\sin\varphi + z\cos\varphi - x\left(a+\frac1a\right)\cos\varphi+\cos\varphi = 0$$
How so?
Well, both equations are certainly true at both points, hence so is any linear combination of them. Our resulting plane passes through both points as well.
How do I know that the equation covers all such planes?
Why, just find the angle between my plane and your second plane, and realize that it is $\varphi$. By varying $\varphi$, you rotate my plane around the line that connects the two points.
So it goes.
Best Answer
The only way a plane can go through exactly $3$ vertices in a cube is through $3$ face diagonals that form a triangle (see the diagram).
The moment a plane includes an edge, it either goes through $2$ vertices or through $4$ vertices. Same is true for a plane going through a space diagonal.
Now the counting:
i) Start with one of the vertices, say vertex $B$. There are $3$ face diagonals through $B$. There are $3$ ways to choose two face diagonals that will uniquely identify a plane.
ii) Then vertex $A$. As $AB$ is an edge, we had no plane counted in (i) that included vertex $A$. So we again have $3$ desired planes through vertex $A$.
iii) Out of $3$ planes through vertex $C$, two of them include diagonal $AC$ so those two are duplicates as they were counted in $(ii)$.
iv) Same with planes through vertex $D$, two of them include diagonal $BD$ so two duplicates, already counted in $(i)$.
Please note that we do not have to count planes through vertices $E, F, G, H$ meeting the criteria as any plane through these vertices meeting the criteria would involve a vertex from the bottom face $ABCD$ which we already counted.
So total number of planes going through exactly $3$ vertices,
$ = 3 + 3 + (3-2) + (3-2) = 8$