Regular Tetrahedron rotation problem

Choose coordinates with the origin at the center of the tetrahedron, and let $a_1,a_2,a_3,a_4\in\mathbb R^3$ be the vertices. Note that $\sum_ia_i=0$. By symmetry $$ a_i\cdot a_j=\begin{cases} \lambda&\text{if }i=j,\\ \mu&\text{if }i\neq j \end{cases} $$ for some $\lambda,\mu$. Choose the scale so that $\lambda=1$ (this means the side lengths are $\sqrt{8/3}$). Then $$ 0=a_1\cdot\sum_i a_i=\lambda+3\mu $$ so $\mu=-1/3$. Also $$\begin{eqnarray*} a_j^T\left(\sum_i a_ia_i^T\right)a_k &=&\sum_i(a_i\cdot a_j)(a_i\cdot a_k)\\ &=&\begin{cases}4/3&\text{if }j=k,\\-4/9&\text{if }j\neq k\end{cases}\\ &=&4/3a_j^Ta_k \end{eqnarray*}$$ Since the $a_i$ span $\mathbb R^3$, we have $\sum_i a_ia_i^T=4/3I$.

Now suppose the plane in question is $\{x\in\mathbb R^3\mid x\cdot n=k\}$ where $|n|=1$. The distance from $a_i$ to the plane is $d_i=a_i\cdot n-k$. Then $$ \sum_i(d_i+k)=\left(\sum_ia_i\right)\cdot n=0, $$ $$ \sum_i(d_i+k)^2=\sum_i n^Ta_ia_i^Tn=4/3|n|^2=4/3. $$ Let $s_1=d_1+d_2+d_3$ and $s_2=d_1^2+d_2^2+d_3^2$. Then $$ s_1+d_4+4k=0, $$ $$ s_2+d_4^2+2k(s_1+d_4)+4k^2=4/3. $$ Combining, $$ 3d_4^2-2s_1d_4+4s_2-16/3-s_1^2=0 $$ Hence $$ d_4=\frac{s_1\pm\sqrt{16-12s_2+4s_1^2}}3. $$ Note that for some values there are two positive solutions, since the tetrahedron can be reflected around the side spanned by $a_1,a_2,a_3$. If you assume the unknown vertex is the highest above the plane, then take the larger solution.