The Hour hand, rotates $360^{\circ}$ in in 12 hours $=12*60$ minutes,
in $t $ minutes rotates $\frac{t}{2}^{\circ}$
The Minute hand, rotates $360^{\circ}$ in $1$ hours=$60$ minutes,
in $t $ minutes rotates $6t^{\circ}$
So, in $t $ minutes, difference of angles between the hands is $6t^{\circ}-\frac{t}{2}^{\circ}$ $=\frac{11t}{2}^{\circ}$
If they make angle $\theta$ (in $^{\circ}$) between them, $\frac{11t}{2}^{\circ}=n360^{\circ}+\theta$, where $n$ is any integer, or $t=\frac{n720^{\circ}+2\theta}{11}$
So, the minimum interval of making angle $\theta$ between them is $\frac{(m+1)720^{\circ}+2\theta}{11}-\frac{m720^{\circ}+2\theta}{11} minute$ $=\frac{720}{11} minute$ (Putting $n=m+1$ and $m$)
So, in $12$ hours, they will make angle $\theta$ between them $\frac{12 hours}{\frac{720}{11} minutes}$ $=11$ times
(1)For coincidence, $\theta=0$
So, in $12$ hours, they will coincide $=11$ times.
(2)For diametrically opposite, $\theta=180^{\circ}$
So, in $12$ hours, they will be diametrically opposite $=11$ times.
(3) To be perpendicular, $\theta=±90^{\circ}$
Clearly, for the '+' sign, there will be 11 occurrences of perpendicularity and so for the '-' sign.
So, in $12$ hours, they will be perpendicular $11+11=22$ times.
Observe that for $180^{\circ}$, we don't consider '±' as $180^{\circ}\equiv -180^{\circ}{\pmod {360^{\circ}}}$