(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. $\square$
