Let $$q=\cos\theta+(x_q\textbf{i}+y_q\textbf{j}+z_q\textbf{k})\sin\theta$$ be a unit quaternion parameterised by $\theta\in\mathbb{R}$, where $(x_q,y_q,z_q)$ is fixed and $x_q^2+y_q^2+z_q^2=1$, and let $$p=w+x\textbf{i}+y\textbf{j}+z\textbf{k}$$ denote another fixed unit quaternion, where $(w,x,y,z)$ is a point on the unit 3-sphere, i.e. $w^2+x^2+y^2+z^2=1$. By identifying $(w,x,y,z)$ with the antipodal point $(-w,-x,-y,-z)$, the quaternion $p$ can be used to represent a point in the 3-dimensional elliptic space.
A left multiplication $$p\mapsto qp$$ can be identified with the left Clifford translation by $\theta$. Allowing $\theta$ to range over $[0,\pi)$, we get a line in the elliptic space (this line maps to a great circle in the 3-sphere), which is Clifford parallel to any other line generated by $p'\mapsto qp'$, where $q$ has the same $(x_q,y_q,z_q)$ as above, $\theta$ ranges over $[0,\pi)$, and $p'=w'+x'\textbf{i}+y'\textbf{j}+z'\textbf{k}$ is an arbitrary unit quaternion. Similar considerations apply to the right Clifford parallels, where a right multiplication is used.
Is there an expression that gives Plücker coordinates of the left and right Clifford parallels passing through $p$ and generated by $q$ as described above? In particular, I'm looking for an expression that perhaps uses quaternionic multiplication and is independent of coordinates as much as possible.