The timing is nearly right ...

At various times of day, the minute and hour hands form a right angle. But is there a time of day when the hour hand, minute hand, and second hand together form two right angles, as illustrated below? If you can find such a time or times, what are they?

If you cannot find any such times, suppose the measures of the two nearly right angles formed by the three hands measure A and B degrees. What time or times of day minimize the square error function $$f(A, B) = (A − 90)^2 + (B − 90)^2?$$

By the rule of the benevolent Fiddler master, we can safely assume that there is no perfect time when the hour, minute and second hands each form right angles with one another. Otherwise, why come up with the second formulation, amirite?

Anywho, we already defined the phase of the hour and minute hands as $h(t)$ and $m(t),$ respectively. Similarly, since the second hand makes 60 full revolutions per hour, its phase can be modeled as $s(t) = 60t - \lfloor 60t \rfloor.$ By taking absolute values of the $\Delta$ function in the Classic problem we get the triangular wave with period $\tau_1 = \frac{12}{11}$ given by $$\Delta_1(t) = | h(t) - m(t) | = \left| \frac{11t}{12} - \left\lfloor \frac{11t}{12} + \frac{1}{2} \right\rfloor \right|.$$ Similarly, we can define the triangular wave with period $\tau_2 = \frac{12}{719}$ given by $$\Delta_2(t) = | h(t) - s(t) | = \left| \frac{719t}{12} - \left\lfloor \frac{719t}{12} + \frac{1}{2} \right\rfloor \right|$$ and the triangular wave with period $\tau_3 = \frac{1}{59}$ given by $$\Delta_3(t) = |m(t) - s(t)| = \left| 59t - \left\lfloor 59t + \frac{1}{2} \right\rfloor \right|.$$

While there is likely some complicated formula for determining which of $\Delta_1,$ $\Delta_2,$ and $\Delta_3$ should be considered the measures A and B, let's ignore this assignment difficult and just reason that we want to let A and B be the minimum and the median of $\Delta_1,$ $\Delta_2$ and $\Delta_3,$ respectively. So we can reframe $f(A,B)$ as $$f(t) = f(\Delta_1, \Delta_2, \Delta_3) = \left( \sum_{i=1}^3 \left(\Delta_i(t) - \frac{1}{4}\right)^2 \right) - \left( \max_i \Delta_i - \frac{1}{4} \right)^2.$$ After an exhaustive search through points in the interval $[0,12],$ our closest almost right times are at $t_1 = 3:49:04.0786$ and $t_2 = 12 - t_1 = 8:10:55.9214,$ each with value of $f(t_1)=f(t_2) = 6.13819\times 10^{-8}.$