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 A_AZ an alias for A_ZS or something else?) and the $\alpha$ on the left side of equation 3.9 is a mistake.
• The solar energy community seem to use $\alpha$ for altitude. In many astronomical contexts where $\delta$ is declination, $\alpha$ is right ascension.
• In that Wikipedia article, the formula for $\cos \phi_s$ with no $h$ term looks wrong.

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 A_AZ an alias for A_ZS 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 $\omega$.

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 $\beta$ = 90$^\circ$. I find their azimuth notation confusing, and the $\alpha$ on the left side of equation 3.9 is a mistake.
• In that Wikipedia article, the formula for $\cos \phi_s$ with no $h$ term looks wrong.

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 A_AZ an alias for A_ZS or something else?) and the $\alpha$ on the left side of equation 3.9 is a mistake.
• The solar energy community seem to use $\alpha$ for altitude. In many astronomical contexts where $\delta$ is declination, $\alpha$ is right ascension.
• In that Wikipedia article, the formula for $\cos \phi_s$ with no $h$ term looks wrong.