Do common sense has anything to do with this question? Because 
corporation might forget the arbitrary 12 airplane rule 
and have 11 times 20 = 220. 
But perhaps they get a discount if all 12 hangars are used! 

Playing by the rules of the question:- 
Let a, b, c be the aeroplanes that carry 
10, 15, and 20 passengers, respectively, 

a + b + c = 12 
10a +15b +20c = 220 

We may rearrange these as 
4a + 4b + 4c = 48 
2a + 3b + 4c = 44 
Now subtract 
2a + b = 4 

There appear to be 3 options 
a = 0, b = 4 which means 220 = 0*10 + 4*15 + 8*20 
a = 1, b = 2 which means 220 = 1*10 + 2*15 + 9*20 
a = 2, b = 0 which means 220 = 2*10 + 0*15 + 7*20 

If you dismiss the ones with zeros it may be that 
the expected answer to “How many of each plane” is 

One that carries 10, 2 that carries 15, and 9 that carries 20