Another question asks for the dihedral angle of a general tetrahedron in terms of its edge lengths. I chose to answer by giving a formulas in terms of face areas (because those relations deserve to be better-known). Most-relevant here is this version, for a tetrahedron $OABC$, ...
$$W^2=X^2+Y^2+Z^2-2YZ\cos\angle OA - 2ZX\cos\angle OB-2XY\cos\angle OC \tag{$\star$}$$
where $W := |\triangle ABC|$, $X := |\triangle OBC|$, $Y := |\triangle AOC|$, $Z := |\triangle ABO|$ are face areas, and $\angle OA$, $\angle OB$, $\angle OC$ are dihedral angles along respective edges $\overline{OA}$, $\overline{OB}$, $\overline{OC}$.
In a regular tetrahedron, all face areas are equal ($W=X=Y=Z$), as are all dihedral angle measures ($\angle OA = \angle OB = \angle OC$), so $$W^2 = 3 W^2 - 6W^2\cos\angle OA \qquad\to\qquad \cos\angle OA = \frac{1}{3} \tag{$\star\star$}$$
I'll leave calculating the angle measure itself to the reader. $\square$
FYI ... The $0$-dimensional point, $1$-dimensional segment, $2$-dimensional triangle and $3$-dimensional tetrahedron have higher-dimensional counterparts, each known simply by the umbrella term "$n$-dimensional simplex" (or "$n$-simplex"). The bounding elements (counterparts of "vertices", "edges", and "faces") are called "facets", and we can write, for $n\geq 2$ ...
Fun Fact. The angle between two neighboring facets of a regular $n$-simplex is $\operatorname{arccos}\frac{1}{n}$ .
So, always remember and never forget: $\cos 60^\circ$ isn't merely "half of one"; it's the reciprocal of the equilateral triangle's dimension!
A more convenient formula for the volume is $$V = \frac{1}{3!} \left|\begin{matrix} x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
x_4 & y_4 & z_4 & 1 \\
\end{matrix}\right|$$
where the vertices are $(x_i, y_i, z_i)$ for $i = 1, 2, 3, 4$ in any order. Take the absolute value of the result.
In WolframAlpha, you can compute this with the input
Abs[Det[{{-0.0865403, -0.122347, 0.898904, 1},
{-0.436523, -0.30131, 1.92251, 1},
{-0.459102, -0.0670386, 1.68168, 1},
{0, 0, 0, 1}}]/3!]
The advantage of this approach is that since one of your vertices is the origin, you could even compute the determinant fairly easily by hand as it reduces to a $3 \times 3$ determinant.
Best Answer
In order to take into account a certain symmetry of this issue, let us consider the intersection line of the planes as the $x$ axis with origin at the midpoint of $p_2p_3$, and the bissector plane as the reference horizontal plane.
Imagine the particular case $\varphi=0$ meaning that the figure is a flat figure in this bissector plane , i.e., all $p_k$ coplanar in the $xy$ plane. From this case, generate the general case by rotating vector $\vec{p_1p_2}$ around the $x$ axis with a $\psi:=\varphi/2$ angle in one direction whereas vector $\vec{p_3p_4}$ is rotated in the other direction i.e., with a $-\psi$ angle.
This will give, using rotation matrices,
$$p_1=\begin{pmatrix}1&0&0\\0&\cos \psi&-\sin \psi\\0&\sin \psi&\ \ \ \ \cos \psi\end{pmatrix}\begin{pmatrix}-a \cos \alpha\\ \color{red}{+}a \sin \alpha\\0 \end{pmatrix}-\begin{pmatrix}b/2\\0\\0\end{pmatrix} $$
$$p_4=\begin{pmatrix}1&0&0\\0&\ \ \ \cos \psi& \sin \psi\\0&-\sin \psi&\cos \psi\end{pmatrix}\begin{pmatrix}-c \cos \beta\\ \color{red}{+}c \sin \beta\\0 \end{pmatrix}+\begin{pmatrix}b/2\\0\\0\end{pmatrix} $$
It remains to expand the previous expressions and then use the distance formula:
$$\delta=\sqrt{(x_1-x_4)^2+(y_1-y_4)^2+(z_1-z_4)^2}$$