Three in one problem

Let the job = 60 units.
Since A can complete the job in 10 days, A's rate = w/t = 60/10 = 6 units per day.
Since B can complete the job in 12 days, B's rate = w/t = 60/12 = 5 units per day.
When A and B work together, their combined rate = 6+5 = 11 units per day.

Case 1: B leaves 2 days before the work is completed
After B leaves, A works for 2 days alone.
Work completed by A in the last 2 days = r*t = 6*2 = 12 units.

Remaining work = 60-12 = 48 units.
The remaining work was completed prior to B's departure and thus was completed by A and B working together.
Time for A and B to complete 48 units = w/r = 48/11 days.

Total time = time for A alone + time for A and B together = 2 + 48/11 = 70/11 days.

Case 2: B leaves 2 days before the scheduled completion
The wording here is unclear.
I assume the scheduled completion is the time it would take A and B to complete the job together.
Time for A and B to complete 60 units = w/r = 60/11 days.

Since B leaves 2 days before the allotted number of days, the time that A and B work together = 60/11 - 2 = 38/11 days.
Work completed by A and B in 38/11 days = r*t = 11 * 38/11 = 38 units.

Remaining work = 60-38 = 22 units.
Time for A to complete the remaining 22 units after B leaves = w/r = 22/6 = 11/3 days.

Total time = time for A and B together + time for A alone = 38/11 + 11/3 = 235/33 days.

Case 3: C, who can destroy the work in 20 days, joins A and B
Thus, C's rate of destruction = w/t = 60/20 = 3 units per day.

Since A and B complete 11 units per day, and C destroys 3 units per day, the combined rate when all 3 work together = 11-3 = 8 units per day.
Time for all 3 working together to complete 60 units = w/r = 60/8 = 15/2 days.