Consider a regular tetrahedron with edge length one (four equilateral triangles joined edge to edge) call it $T$. Set $T$ on the $x,y$-plane with a vertex at the origin and an edge aligned with the positive $x$-axis. Call the three edges that touch the origin $i$, $f$, and $g$, where $i = (1, 0, 0)$, $f = \left(\frac12, \frac{\sqrt3}{2}, 0\right)$.
Find the equation for the plane that contains the face of $T$ not touching the origin.
Step one is to find $g$, which I am getting as $\left(\frac12, \frac{1}{2\sqrt3}, \sqrt{\frac23}\right)$. This is confirmed by the answer key I am working with.
Now, to find the plane I subtracted both $i$ from both $g$ and $f$, getting me vectors, I then took the cross product of those and that would be my $A, B, C$ of the plane equation. Then I took the dot product of that vector with $i (1, 0, 0)$ and got $\frac{\sqrt6}{2\sqrt3}$ as my $D$ value, this left me with an answer of:
$$\frac{\sqrt2}{2} x + \frac{\sqrt2}{2\sqrt3} y + \frac{1}{2\sqrt3} z + \frac{\sqrt6}{2\sqrt3} = 0.$$
This however, is not what my answer key has. Where did I go wrong?
If you take the sum of $i, f, g$, it will be normal to the plane you want by symmetry. So the normal is parallel to $(2, 2 \sqrt{3}/3, \sqrt{2/3})$.
Then your equation for the plane is
$$2x + 2\sqrt{3}y/3 + \sqrt{2}z/\sqrt{3} - 2= 0,$$
where we used point-slope form with the point as $(1,0,0)$.
I'm not sure what your answer key says, but this equation satisfies all three points you found, so it seems to check out.
Edit: Good news! This is the same equation as the answer key:
$$x + y/\sqrt{3} + z/\sqrt{6} - 1 = 0$$
Divide mine by $2$ on both sides and you'll get the answer key's answer.