In the below diagram, the angle $\gamma$ represents the dihedral angle between the plane formed by $r_1R$ and the plane formed by $r_2R$. The origin is at the intersection between $r_1$ and $R$.
I know that the $x$ and $y$ coordinates for points $A$, $B$, $C$, and $D$ have $\cos(\alpha)$, $\sin(\alpha)$, $\cos(\beta)$, and $\sin(\beta)$ terms, but I am unsure how to incorporate the dihedral angle, $\gamma$, for $x$, $y$, and $z$.
To get $C$ and $D$ from $A$ and $B$, assume $\vec{A} = -\vec{B} = -\frac{r_1}{2}\begin{bmatrix}\sin(\alpha)\\ \cos(\alpha)\\0\end{bmatrix}$.
(1) Rotate the Line $AB$ around the $x$-axis $\gamma$ degrees.
Rotating $\vec{A}$ (the line from the origin $0$ to the point $A$) by an angle of $\gamma$ around the $x$-axis yields $$\vec{A'}=-\frac{r_1}{2}\begin{bmatrix}\sin(\alpha)\\ \cos(\alpha)\cos(\gamma)\\\cos(\alpha)\sin(\gamma)\end{bmatrix}, $$ now the line between $0$ and $A'$ is parallel to the line $CD$.
(2) Change the size of $r_1$ to $r_2$ or scale $r_1$ by $\frac{r_2}{r_1}$.
Scaling $\vec{A'}$ with $\frac{r_2}{r_1}$ yields $$\vec{A''}=-\frac{r_2}{2}\begin{bmatrix}\sin(\alpha)\\ \cos(\alpha)\cos(\gamma)\\\cos(\alpha)\sin(\gamma)\end{bmatrix}.$$
(3) Move (the rotated and scaled line $AB$) exactly $R$ units along the positive $x-$direction
Adding $R$ units along the positive $x-$direction to $\vec{A''}$ we get
$$\vec{C}=-\vec{D}=-\frac{r_2}{2}\begin{bmatrix}\sin(\alpha)-\frac{2}{r_2}R\\ \cos(\alpha)\cos(\gamma)\\\cos(\alpha)\sin(\gamma)\end{bmatrix}. $$
EDIT
$r(x)=r_1 + (r_2-r_1)\cdot\left(\frac{x}{R}\right)$