The triangle circumcenter is conveniently expressed in trilinear coordinates as $$ \cos\alpha_1 : \cos\alpha_2 : \cos\alpha_3, $$ where the $\alpha_i$ are the angles opposite of the respective edge in the trilinear system.
Is there a similar expression for the circumcenter in quadriplanar coordinates, i.e., the same concept carried over to tetrahedra?
Let's consider the tetrahedron
and let's say we'd like to find out about the quadriplanar coordinate corresponding to point $A$. Let's further assume that the edges are directed like this: $$ a' = \overrightarrow{AD}, b = \overrightarrow{AC}, c = \overrightarrow{AB}\\ b' = \overrightarrow{BD}, c' = \overrightarrow{DC}, a = \overrightarrow{CB}. $$ Then the elevation of the circumcenter above the plane spanned by $C, B, D$ is given by $$ c_A = \\ (\langle a, b' \rangle \langle b', c' \rangle \langle c, a' \rangle +\\ \langle b', c' \rangle \langle c', a \rangle \langle a', b \rangle +\\ \langle c', a \rangle \langle a, b' \rangle \langle b, c \rangle +\\ \langle a, b' \rangle \langle b', c' \rangle \langle c', a \rangle )\\ / 24\\ /\alpha_{BDC}\\ / v, $$ where $v$ is the volume of the tetrahedron and $\alpha_{BDC}$ the area of the face $BDC$.
The same can be done for all other faces.
Since the quadriplanar coordinates are invariant under multiplication, the fixed factors $24$ and $v$ can be left aside.