I think there is a much simpler solution than all provided so far.
Consider the following three facts:
- hour hand will rotate 2 full times in a day
- minute hand will rotate 24 full times in a day
- second hand will rotate 1440 full times in a day
So then:
(a) the minute and hour hands, will meet exactly:
22 times: 24 - 2 (once every 24/22 hours)
(b) the minute and second hands, will meet exactly:
1416 times: 1440 - 24 (once every 24/1416 hours)
(c) the hour and second hands, will meet exactly:
1438 times: 1440 - 2 (once every 24/1438 hours)
(d) all three hands, will meet exactly:
twice: only at exactly 12:00:00 o'clock (noon and midnight)
Simply because:
The faster hand passes the slower hand by the number of laps it makes minus the number of laps the slower hand makes.
With the special case:
One hand lapping another won't necessarily coincide with the third hand being there. Try to find the common multiples of the three fractions 24/22, 24/1416, and 24/1438 and you will see there are only two at 24/1 and 24/2. (ie. 12 hours and 24 hours after the start).