For opposing edges in a tetrahedron, define $p\otimes q = p^2 + q^2 + 2 p q \cot(\angle p)\cot(\angle q)$, where $\angle p$ is the inner dihedral angle of edge $p$.
In tetrahedron ABCD, $AB\otimes CD = AC\otimes BD =AD\otimes BC$, a property of dihedral angles in a tetrahedron. I'll call this the dihedral constant of a tetrahedron.
Given tetrahedron ABCD, does there exist a point E such that the dihedral constants of ABCE, ABDE, ACDE, and BCDE are identical? It turns out the point does exist.
For $((0,0,0),(\sqrt2,0,0),(0,\sqrt3,0),(0,0,\sqrt6))$, $(0.2925509218717602,0.28324348233978136,0.32488931587547265)$ works
For $((0,0,0),(5,0,0),(6/5,(12/5) \sqrt6,0),(-5,0,4 \sqrt6)$, $(2.406010889675711,5.878903865331654,0.877729911045533)) $ works
For $((0,0,0),(5,0,0),(19/5,(12 \sqrt6)/5,0),(-(31/5),(12 \sqrt6)/5,4 \sqrt6)$, $(-0.2458533044423744,6.927220621203815,-3.3113164490462124))$ works.
The dihedral constant center is off of the plane of the usual tetrahedron centers. In the image below the black spot is the dihedral constant center, corresponding to the third solution above.
For any of these three tetrahedra, can anyone find an exact solution, or a general exact solution?
Best Answer
(My labeling doesn't match OP's.)
Consider tetrahedron $OABC$ with faces (and face-areas) $W$, $X$, $Y$, $Z$ opposite vertices $O$, $A$, $B$, $C$. Define $$a := |OA| \quad b := |OB| \quad c := |OC| \quad d := |BC| \quad e := |CA| \quad f := |AB|$$ and let $V$ be the volume. Also, define dihedral angles $A$, $B$, $C$, $D$, $E$, $F$ along the edges with corresponding lower-case labels. (There should be no confusion with using "$A$" for both a vertex and an angle.) First, the "dihedral constant" is given by the formula I posted in a comment:
$$\delta(OABC) = \frac{1}{9V^2}\left(\begin{array}{c} -W^4-X^4-Y^4-Z^4 +2W^2X^2+2W^2Y^2+2W^2Z^2 \\ +2Y^2Z^2+2Z^2X^2+2X^2Y^2 \end{array}\right) $$
Consider a point defined by the coordinate-vector equation $$P := p\,A + q\,B + r\,C + s\,O \qquad\text{where}\quad p + q + r + s = 1$$ We'll see that $p$, $q$, $r$, $s$ become closely associated with respective opposite faces $X$, $Y$, $Z$, $W$.
It's possible to write the dihedral constants of the tetrahedrons determined by $P$ in terms of the elements of the original tetrahedron. Pardon a bit more notation, but ... To reduce some visual clutter in the formulas, we define $m^2 = \delta(OABC)$, as well as $W_s :=W/s$, $X_p := X/p$, $Y_q := Y/q$, $Z_r:= Z/r$ and $$t_A := Y_q Z_r \cos A \qquad t_B := Z_r X_p \cos B \qquad t_C := X_p Y_q \cos C$$ $$t_D := W_s X_p \cos D \qquad t_E := W_s Y_q \cos E \qquad t_F := W_s Z_r \cos F$$ With these, we have ...
$$\begin{align} \delta(PABC) = &- \left(\;p s\,a^2 + q s\,b^2 + r s\,c^2 + q r\,d^2 + p r\,e^2 + p q\,f^2 \;\right) \\[4pt] &+ p\,d^2 + q\,e^2 + r\,f^2 + s\,m^2\\[4pt] &+ \frac{8 W^2\,p q r}{9V^2} \left(\;-W_s^2 + (\;t_A + t_B + t_C\;) - (\;t_D + t_E + t_F\;)\;\right) \end{align}$$
To make some sense of the alphabet soup, first observe that tetrahedrons $OABC$ and $PABC$ have face $W$ in common. Now,
In the first grouping of terms, $a$ is the edge between faces $Y$ and $Z$, hence it's opposite the edge ($d$) between $W$ and $X$. The constants associated with the opposite-edge faces are $s$ and $p$, which we see in the term $ps\,a^2$. Likewise, $b$ is opposite the edge between faces $W$ and $Y$, which are associated with $s$ and $q$, and we have the term $qs\,b^2$. Etc. This grouping is symmetric in the elements of the tetrahedron; we'll see it again.
In the second grouping, edges $d$, $e$, $f$ surround face $W$. Constants $p$, $q$, $r$ are associated with the respective faces ($X$, $Y$, $Z$) adjacent to $W$ across those edges. The left-over constant, $s$, goes with the dihedral constant of the original tetrahedron.
In the third grouping, the multiplied constant features the common face $W$ and the product ($pqr$) of constants except the one associated with that face. For the $t$-terms, observe that $D$, $E$, $F$ are the dihedral angles along the edges surrounding $W$, while $A$, $B$, $C$ are the angles surrounding opposite vertex $O$.
The reader can double-check those rules (and my typing) by comparing the expressions for the other constants:
$$\begin{align} \delta(OPBC) = &-\left(\;p s\,a^2 + q s\,b^2 + r s\,c^2 + q r\,d^2 + p r\,e^2 + p q\,f^2 \;\right) \\[4pt] &+ p\,m^2 + q\,c^2 + r\,b^2 + s\,d^2 \\[4pt] &+ \frac{8 X^2\,q r s}{9V^2} \left(\;-X_p^2 + (\;t_A + t_E + t_F\;) - (\;t_D + t_B + t_C \;)\;\right) \\[8pt] \delta(OAPC) = &-\left(\;p s\,a^2 + q s\,b^2 + r s\,c^2 + q r\,d^2 + p r\,e^2 + p q\,f^2 \;\right) \\[4pt] &+ p\,c^2 + q\,m^2 + r\,a^2 + s\,e^2 \\[4pt] &+ \frac{8 Y^2\,p r s}{9V^2} \left(\;-Y_q^2 + (\; t_D + t_B + t_F \;) - (\; t_A + t_E + t_C \;) \;\right) \\[8pt] \delta(OABP) = &-\left(\;p s\,a^2 + q s\,b^2 + r s\,c^2 + q r\,d^2 + p r\,e^2 + p q\,f^2 \;\right) \\[4pt] &+ p\,b^2 + q\,a^2 + r\,m^2 + s\,f^2 \\[4pt] &+ \frac{8 Z^2\,p q s}{9V^2} \left(\;-Z_r^2 + (\;t_D + t_E + t_C \;) - (\; t_A + t_B + t_F \;)\;\right) \end{align}$$
(As promised, the symmetric first grouping appears in all the formulas.)
In any case ... The search for a Dihedral Constant Point reduces to solving for $p$, $q$, $r$ (and $s=1-p-q-r$) such that $$\delta(PABC) = \delta(OPBC) = \delta(OAPC) = \delta(OABP) \tag{$\star$}$$
This turns out to be no easy feat. Even in the case of OP's right-corner tetrahedron $O=(0,0,0)$, $A=(\sqrt{2},0,0)$, $B=(0,\sqrt{3},0)$, $C=(0,0,\sqrt{6})$, eliminating, say, $q$ and $r$ (and $s$), from system $(\star)$ leaves an irreducible degree-$27$(!) polynomial in $p$. (I'm ignoring some extraneous factors that Mathematica is showing me.) Surprisingly, the polynomial has a single real root corresponding to OP's solution. It seems unlikely that there's a closed form for this value.
I won't carry out the full analysis here, but I'll show how the formulas for the Dihedral Constant simplify in the case of a right corner tetrahedron $OABC$ with hypotenuse-face $W$. From the right triangular faces, we have $$d^2 = b^2 + c^2 \qquad e^2 = c^2 + a^2 \qquad f^2 = a^2 + b^2 \qquad X = \frac12 b c \qquad Y = \frac12 ca \qquad Z = \frac12 ab$$ Also, $$\cos A = \cos B = \cos C = 0 \quad\to\quad t_A = t_B = t_C = 0$$ $$\cos D = \frac{X}{W} \quad \cos E = \frac{Y}{W} \quad \cos F = \frac{Z}{W} \quad\to\quad t_D = \frac{1}{ps}X^2 \quad t_E = \frac{1}{qs}Y^2 \quad t_F = \frac{1}{rs}Z^2$$ $$W^2 = X^2 + Y^2 + Z^2 \qquad V = \frac16 a b c \qquad m^2 = \delta(OABC) = a^2 + b^2 + c^2$$
Making appropriate substitutions and manipulations, we achieve
$$\begin{align} \delta(PABC) = &\phantom{+}k^2 - 2 \left( p\,a^2 + q\,b^2 + r\,c^2 \right) \\[4pt] &- \frac{2 \left( a^2 b^2 + b^2 c^2 + c^2 a^2 \right)}{s^2\,a^2b^2c^2} \left(\; q r\,b^2 c^2 ( p + s ) + p r\,c^2 a^2 ( q + s ) + p q\,a^2 b^2 ( r + s ) \;\right)\\[8pt] \delta(OPBC) = &\phantom{+}k^2 - a^2 - \frac{2 q r \,b^2 c^2(p+s)}{p^2\,a^2}\\[8pt] \delta(OAPC) = &\phantom{+}k^2 - b^2 - \frac{2 p r \,a^2 c^2(q+s)}{q^2\,b^2} \\[8pt] \delta(OABP) = &\phantom{+}k^2 - c^2 - \frac{2 p q \,a^2 b^2(r+s)}{r^2\,c^2} \end{align}$$ where $k^2 := a^2 ( 1 + p^2 ) + b^2 ( 1 + q^2 ) + c^2 ( 1 + r^2 )$ is a common summand that cancels (so, can be ignored) in $(\star)$. The reader (in particular, OP, who is fluent in Mathematica) is invited to verify that, when $a=\sqrt{2}$, $b=\sqrt{3}$, $c = \sqrt{6}$, the system has solution $$(p,q,r) = (0.20686\ldots, 0.16353\ldots, 0.13263\ldots)$$ which corresponds to Dihedral Constant Point $(pa, qb, rc)$ as given by OP. I'm also leaving consideration of the non-right tetrahedral examples to the reader.
Addendum. It may be worth noting that I derived the face-based formula for the Dihedral Constant by invoking some basic hedronometric relations. Specifically,
$$Y^2 + Z^2 - 2 Y Z \cos A \;=\; H^2 \;=\; W^2 + X^2 - 2 W X \cos D$$ $$[H,Y,Z] = 4 Y^2 Z^2 \sin^2 A = 9 V^2 a^2 \qquad [H,W,Z] = 4 W^2 Z^2 \sin^2 D = 9 V^2 d^2$$
Here, $H$ is what I call a pseudoface. It's essentially the quadrilateral shadow of the tetrahedron cast on a plane parallel to a pair of opposite edges ($a$ and $d$ in the case of pseudoface $H$), but one can define it formally via the cosine relation. The sine relation involves the "Heronic product": $$[x,y,z]:=(x+y+z)(-x+y+z)(x-y+z)(x+y-z)$$ (so named because of its use in Heron's formula for the area of a triangle). The relations imply, for instance, that $$a \cot A = a\;\frac{\cos A}{\sin A} = \frac{\sqrt{[H,Y,Z]}}{3V}\;\frac{\left(Y^2+Z^2-H^2\right)/(2YZ)}{\sqrt{[H,Y,Z]}/(2YZ)} = \frac{Y^2+Z^2-H^2}{3V}$$ Thus, $$\begin{align} a \otimes d &:= \frac{1}{9V^2}\left(\;[H,Y,Z] + [H,W,X] + 2\,\left(Y^2+Z^2-H^2\right)\left(W^2+X^2-H^2\right)\;\right) \end{align}$$ which expands into the formula given above. $\square$