A regular tetrahedron is centered at the origin with vertices $ (0,0,1$) and $(a,0,b)$. All of its vertices satisfy $x^2+y^2+z^2=1$
Find the remaining two vertices with respect to $a$ and $b$.

To solve everything from scratch will result in a too long answer,
so we assume these important properties to be known:

Image 1: Tetraeder in cube construction. (Source)


*

*The tetraeder in the task has the internal arm length $1$, meaning
the distance from origin to top vertex. Then the edge length between
two vertices is $L=2\sqrt{2/3}$.
(This corresponds to a cube length $s=2/\sqrt{3}$ in the above image)

*The internal angle between two arms is $\theta\approx 109^0$ with
$\arccos \theta = -1/3$.
(This is only used to check the following results)


For the second vertex $Q=(a,0,b)$ we have the constraints that it lies on the unit sphere around the origin and on the sphere around $P=(0,0,1)$ of radius $L$.
\begin{align}
1 &= x^2 + y^2 + z^2 \\
L^2 &= x^2 + y^2 + (z-1)^2
\end{align}
The intersection is the circle at height
$$
L^2 = 2-2 z \iff z = 1 - L^2/2 = 1-4/3 = -1/3
$$
with the equation
$$
x^2+y^2=1-1/9=(2\sqrt{2}/3)^2 \quad (*)
$$
which has radius $r = 2 \sqrt{2}/3$.

(Large version)
Image 2: Unit sphere constraint (red), $L$ sphere constraint (green). plane constraints $z=-1/3$ and $y = 0$.
Cut with the plane $y=0$ this leaves two choices:
$$
Q=\left( \pm2\sqrt{2}/3,0,-1/3 \right)
$$
Note that we can pick only one of these solutions and not both as feasible vertices, as the angle between them is not $\theta$. 
We continue with the positive choice $Q_1$.
The constraints for $P$ and $A$ again lead to two choices for the third vertex $R$:


(Large version left image) (Large version right image)
Images 3 and 4: Unit sphere (purple) and $L$ sphere constraints for $P$ (red) and $Q$ (green). Intersection circles (yellow).
The intersection of unit sphere and $L$ sphere of $Q$
$$
1 = x^2 + y^2 +z^2 \\
L^2 = (x-2\sqrt{2}/3)^2 + y^2 + (z+1/3)^2
$$
gives
$$
L^2 = 8/3 = 1 -(4\sqrt{2}/3)x + 8/9 + (2/3)z + 1/9 \iff \\
1 = - 2\sqrt{2}x + z
$$
as equation of the plane where the intersection circle lies in.
Intersecting this plane with circle $(*)$ we get the equations
$$
1 = -2\sqrt{2} x -1/3 \iff
x = -\sqrt{2}/3
$$
and
$$
8/9 = 2/9 + y^2 \iff y = \pm \sqrt{6}/3
$$
which gives two feasible vertices
$$
R = (-\sqrt{2}/3, \pm \sqrt{6}/3, -1/3)
$$
It turns out that these are the remaing two vertices we are looking for.

(Large version)
Image 5: The resulting tetraeder $(P,Q_1, R_1, R_2)$. Note the two measured tetraeder angles $\theta$. Note the alternative tetraeder vertex $Q_2$.
Summary:
We found two tetraeder that fit the task.
$$
\begin{array}{l}
P = \left( 0, 0, 1 \right) \\
Q_1 = \left( 2\sqrt{2}/3,0,-1/3 \right) \\
R_! = \left( -\sqrt{2}/3, \sqrt{6}/3, -1/3 \right) \\
R_2 = \left( -\sqrt{2}/3, -\sqrt{6}/3, -1/3 \right)
\end{array}
\quad\quad
\begin{array}{l}
P = \left( 0, 0, 1 \right) \\
Q_2 = \left( -2\sqrt{2}/3,0,-1/3 \right) \\
R_3 = \left( \sqrt{2}/3, \sqrt{6}/3, -1/3 \right) \\
R_4 = \left( \sqrt{2}/3, -\sqrt{6}/3, -1/3 \right)
\end{array}
$$
