Tetrahedron Centers
For the comparatively-easy ones ...
As in my other answer, given tetrahedron $PABC$, I use barycentric coordinates $\rho$, $\alpha$, $\beta$, $\gamma$ to parameterize the point of interest as $$\frac{\rho\,P + \alpha\,A + \beta\, B + \gamma\,C}{\rho + \alpha + \beta + \gamma}$$
We'll take $V$ to be the volume of the tetrahedron. Other elements (side-lengths, face-areas, dihedral angles) are as labeled in the figure:
Of note:
- Opposite edge-pairs are $(a,d)$, $(b,e)$, $(c,f)$.
- Edges surrounding vertices $P$, $A$, $B$, $C$ are $(a,b,c)$, $(a,e,f)$, $(d,b,f)$, $(d,e,c)$.
- Edges surrounding faces $W$, $X$, $Y$, $Z$ are $(d,e,f)$, $(d,b,c)$, $(a,e,c)$, $(a,b,f)$.
See the addendum to this recent answer about how to convert edge-length expressions into hedronometric (face-area-based) forms, which may be more appropriate in this context. (Finding the "best" hedronometric form is something of a challenge.)
Centroid ($G$)
$$\rho : \alpha : \beta : \gamma \;=\; 1 : 1 : 1 : 1$$
Incenter ($I$)
$$\rho : \alpha : \beta : \gamma \;=\; W : X : Y : Z$$
Circumcenter ($O$)
$$\begin{align}\rho &=\phantom{+} a^2 d^2 \left(-d^2 + e^2 + f^2 \right) \\ &\phantom{=}+ b^2 e^2 \left(\phantom{-}d^2 - e^2 + f^2 \right) \\[4pt] &\phantom{=}+ c^2 f^2 \left(\phantom{-}d^2 + e^2 - f^2 \right) \\[4pt] &\phantom{=}- 2 d^2 e^2 f^2 \\[8pt] &= 18 V^2 - a^2\,W X \cos D - b^2\,W Y \cos E - c^2\,W Z \cos F \end{align}$$
Nine/Twelve-Point Center ($T$)
(I'm taking this to mean the center of sphere through the centroids of the faces, which also contains other points of interest, as described in the Wikipedia "Tetrahedron" entry.)
$$\begin{align}\rho \;=\; &\phantom{-\;} 2 a^2 d^2 \left(-a^2 + b^2 + c^2 \right) + a^2 d^2 \left(-d^2 + e^2 + f^2 \right) \\[4pt] &+ 2 b^2 e^2 \left(\phantom{-}a^2 - b^2 + c^2 \right) + b^2 e^2 \left(\phantom{-}d^2 - e^2 + f^2 \right) \\[4pt] &+ 2 c^2 f^2 \left(\phantom{-}a^2 + b^2 - c^2 \right) + c^2 f^2 \left(\phantom{-}d^2 + e^2 - f^2 \right) \\[4pt] &- 2 b^2 c^2 d^2 - 2 a^2 c^2 e^2 - 2 a^2 b^2 f^2 \end{align}$$
Isogonal Conjugates, and the Symmedian Point ($K$)
According to the Forum Geometricorum article "Isogonal Conjugates in a Tetrahedron" (PDF) by Sadek, et al., if isogonal conjugates have barycentric coordinates $(\rho,\alpha,\beta,\gamma)$ and $(\rho^\prime, \alpha^\prime, \beta^\prime, \gamma^\prime)$, then
$$\frac{\rho\rho^\prime}{W^2} = \frac{\alpha\alpha^\prime}{X^2} = \frac{\beta\beta^\prime}{Y^2} = \frac{\gamma\gamma^\prime}{Z^2}$$
(This result reconfirms that the Incenter ($I$) is its own isogonal conjugate. But I digress ...) Thus, $$\rho^\prime : \alpha^\prime : \beta^\prime : \gamma^\prime \;=\; \frac{W^2}{\rho} : \frac{X^2}{\alpha} = \frac{Y^2}{\beta} = \frac{Z^2}{\gamma}$$
Since the Symmedian Point ($K$) is the isogonal conjugate of the centroid ($G$), its barycentric coordinates satisfy
$$\rho : \alpha : \beta : \gamma \;=\; W^2 : X^2 : Y^2 : Z^2$$
Monge Point ($M$)
$$\begin{align} \rho &=\phantom{+}a^2d^2 \left(-a^2 + b^2 + c^2\right) \\[4pt] &\phantom{=\,}+ b^2 e^2 \left(\phantom{-}a^2 - b^2 + c^2\right) \\[4pt] &\phantom{=\,}+ c^2 f^2 \left(\phantom{-}a^2 + b^2 - c^2\right) \\[4pt] &\phantom{=\,}+ d^2 e^2 f^2 - b^2 c^2 d^2 - a^2 c^2 e^2 - a^2 b^2 f^2 \end{align}$$
Spieker Point ($S$)
The Spieker Point is the incenter of the medial tetrahedron (ie, the tetrahedron whose vertices are the centroids of the faces) of $PABC$.
By the Incenter formula above, $S$ is given by
$$\frac{W^\prime\,P^\prime + X^\prime\,A^\prime + Y^\prime\,B^\prime + Z^\prime\,C^\prime}{W^\prime + X^\prime + Y^\prime + Z^\prime}$$
for the medial tetrahedron with vertices $P^\prime = \frac13(A+B+C)$, etc, and face-areas $W^\prime = \frac19 W$, etc. This can be re-written as
$$\frac{\frac13(X+Y+Z)\,P + \frac13(W+Y+Z)\,A + \frac13(W+X+Z)\,B + \frac13(W+X+Y)\,C}{W + X + Y + Z}$$ from which we observe $$\rho : \alpha : \beta : \gamma \;=\; X + Y + Z : W + Y + Z : W + X + Z : W + Z + Y $$
Orthocenter ($H$)
In general, a tetrahedron does not have an orthocenter. Equivalent conditions that force a tetrahedron to be orthocentric are $$a^2 + d^2 = b^2 + e^2 = c^2 + f^2 $$ $$\cos A \cos D = \cos B \cos E = \cos C \cos F$$ $$\overrightarrow{PA}\perp\overrightarrow{BC} \qquad \overrightarrow{PB}\perp\overrightarrow{CA} \qquad \overrightarrow{PC}\perp\overrightarrow{AB}$$
Note: If any two orthogonality conditions hold, then the third one rides for free.
Symmetric-looking barycentric coordinates for the orthocenter of such a tetrahedron are a bit elusive. I'll have to come back to this.
Inspired by this previous question by OP, I investigated "hedronometric" (face-area-based) counterparts of a triangle's Congruent Isoscelizers Point and Congruent Parallelians Point. I wrote a note about them.
In the following, barycentric coordinates $\rho$, $\alpha$, $\beta$, $\gamma$ describe the point $$\frac{\rho\,P + \alpha\,A + \beta\,B + \gamma\,C}{\rho + \alpha + \beta + \gamma} \tag{$\star$}$$ The corresponding trilinear (er, um, tetraplanar) coordinates of the point derive by dividing each barycentric by the area of the appropriate opposite face (here, $W$, $X$, $Y$, $Z$):
$$\frac{\rho}{W} : \frac{\alpha}{X} : \frac{\beta}{Y} : \frac{\gamma}{Z} \tag{$\star\star$}$$
Equal-Area Parallelians Point
Definition and Theorem. A parallelian of tetrahedron $PABC$ with respect to vertex $P$ is a triangle, with vertices on the edge-lines through $P$, in a plane parallel to that of $\triangle ABC$. Any non-degenerate tetrahedron admits a unique point common to the planes of four equal-area parallelians (one for each vertex); this is the Equal-Area Parallelians Point (EAPP).
The barycentric coordinates of the EAPP are $$\rho =-\frac{2}{\sqrt{W}}+\frac{1}{\sqrt{X}}+\frac{1}{\sqrt{Y}}+\frac{1}{\sqrt{Z}}\qquad \alpha =\phantom{-}\frac{1}{\sqrt{W}}-\frac{2}{\sqrt{X}}+\frac{1}{\sqrt{Y}}+\frac{1}{\sqrt{Z}}\qquad\text{etc} \tag{1}$$
Equal-Area Trisohedralizers Point
Definition and Theorem. A trisohedralizer of a tetrahedron $PABC$ with respect to vertex $P$ is a triangle $\triangle A^\prime B^\prime C^\prime$, with vertices on the edge-lines through $P$, that serves as a base for a "trisohedral" tetrahedron. (That is, $|\triangle PB^\prime C^\prime| = |\triangle P C^\prime A^\prime| = |\triangle P A^\prime B^\prime |$.) Any non-degenerate triangle admits a unique point common to the planes of four equal-area trisohedralizers (one for each vertex); this is the Equal-Area Trisohedralizers Point (EATP).
The barycentric coordinates of the EATP are $$\rho = W \left(\;2\sqrt{W/\lambda_P} - \sqrt{X/\lambda_A} - \sqrt{Y/\lambda_B} - \sqrt{Z/\lambda_C}\;\right) \qquad \text{etc} \tag{2}$$ where $$\lambda_V^2 \;=\; 3 - 2\;\sum_{\theta} \cos\theta \tag{3}$$ with the sum taken over the three dihedral angles along edges emanating from vertex $V$.