= Solution
Use column vectors in a right-handed frame, and interpret clockwise as viewed from the positive $x$-axis towards the origin. The <rotation matrix> fixes $e_x$ and sends $e_y$ towards $-e_z$. With $q=1/\sqrt2$, its images of the basis vectors give
$$
\boxed{R=\begin{pmatrix}1&0&0\\0&q&q\\0&-q&q\end{pmatrix}.}
$$
The transverse block is the <planar rotation> through $-\pi/4$. Viewing from the other end of the axis reverses the meaning of clockwise and replaces $R$ by $R^T$; the viewpoint is not specified in the paper.
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