Angles of sunrise and sunset viewed from a surface anyway oriented
In other words, at what hour angle $h$ is the Sun's azimuth $\mathtt{Az}_s$ at a right angle to the wall's normal vector (84$^\circ$ or 264$^\circ$)?

Using these formulas, which I've checked, from
Wikipedia: Solar azimuth angle,
$$\sin \phi_s = \frac{-\sin h \cos \delta}{\sin \theta_s} \\
\cos \phi_s = \frac{\sin \delta \cos \Phi - \cos h \cos \delta \sin \Phi}{\sin \theta_s}$$
where $\phi_s$ is the Sun's north-clockwise azimuth,
$\theta_s$ is the Sun's zenith angle (90$^\circ$ - $\mathtt{Alt}_s$),
$\delta$ is the Sun's declination,
and $\Phi$ is your geographic latitude.
Then $\tan \phi_s = \sin \phi_s / \cos \phi_s$, and
$$\phi_s = \mathtt{atan2}(-\sin h \cos \delta, \ \sin \delta \cos \Phi - \cos h \cos \delta \sin \Phi)$$
This formula yields results consistent with the
NOAA Solar Calculator in my tests.
We need to solve for $h$, but how isn't yet obvious to me; maybe someone else can help?

Notes:


The relation between $\phi_s$ and $\mathtt{Az}_s$ depends on whether $\mathtt{Az}_s$ is south-clockwise, south-counterclockwise, or the more conventional north-clockwise.
$h$ here is the same as $\omega$ in the ITACA article.
As you observed, formula 3.7 is undefined for a wall facing due north or south.
I find their azimuth notation confusing (is AAZ an alias for AZS or something else?) and the $\alpha$ on the left side of equation 3.9 is a mistake.
I believe ITACA formula 3.11 is correct. If we let $\cos \theta_i$ = 0, it reduces to something like the above where the surface is vertical, but it has the same difficulty of solving for $\omega$.