Solution:
Let t∈[0,2] represent the position of the hour hand, i.e., how many full revolutions it has made. Then, the position of the minute hand is 12t (it makes 12 full revolutions per 1 revolution of the hour hand), and the position of the second hand is 720t (it makes 60 full revolutions per 1 revolution of the minute hand). Then, in order for equal angles to be formed, we need (a−b)−(b−c)=a−2b+c≡0(mod1), where a,b,c is a permutation of t,12t,720t. (Here, b would correspond to the hand that's the angle bisector.) Checking all three possibilities,
12t−2(t)+720tt−2(12t)+720tt−2(720t)+12t≡697t≡0(mod1)≡730t≡0(mod1)≡−1427t≡0(mod1)
Then we require t to be a multiple of 6971, 7301, or 14271. Since 697, 730, and 1427 are pairwise relatively prime, the possible values of t are
6971,6972,…,697696,697698,…,6972⋅697−1,7301,7302,…,730729,730731,…,7302⋅730−1,14271,14272,…,14271426,14271428,…,14272⋅1427−1
since t∈[0,2]. This gives a count of 2((697−1)+(730−1)+(1427−1))=5702.
Note that in the above count we don't count t=0,1,2 since then all three hands would overlap. If two hands overlap, then one of 11t,708t,719t≡0(mod1), and the only way one of these can happen and t being a multiple of 6971, 7301, or 14271 is if t=21 and t=23 (which correspond to 6:00 AM and PM). This is because the only pair of numbers that are not relatively prime among 11, 708, 719, 697, 730, 1427 is 708 and 730. The only common divisor of these two numbers is 2, hence t=21,23. Thus the final answer is 5702−2=5700.