Can you read the time?

Let the angular speeds of hour-hand and minute-hand by $\omega_h$ and $\omega_m$ respectively, then we have $\omega_h = \dfrac {30^\circ}{60} = \frac 12 ^\circ \text{/min}$ and $\omega_m = \dfrac {360^\circ}{60} = 6^\circ \text{/min}$ .

At 10:00 o'clock, $\angle AOT = 60^\circ$ and $\angle BOT = 0^\circ$ . Let the required $\angle BOT-\angle AOT$ be $\theta$ and the time after 10:00 to reach the required angles be $t$ minutes, then for minute-hand: $\theta = \omega_m t = 0.5 t$ and hour-hand: $\theta = 60 - \omega_h t = 60 - 6t$ , then:

$\begin{aligned} \frac t2 & = 60-6t \\ t & = 120 - 12t \\ 13 t & = 120 \\ \implies t & = \frac {120}{13} \text{ min} \approx 9 \text{ min } 14 \text{ s} \end{aligned}$

The time is therefore $10:9:14$ , $\implies X+Y+Z = 10+9+14 = \boxed{33}$