The positive integer are 2,3,5 and 7 and this has a common difference of 2 without considering the first term which the the only number given in the equation above.
Here is the solution.
Where a = 2 , b = 3 and c = 5
a! + b! + c! = (21)+(321)+(5432*1)
a! + b! + c! = (2) + (6) + (120)
a! + b! + c! = 128
From equation 2^n = 128 where 128 can be converted to power of 2 which is 2^7
Therefore n = 7, a = 2, b = 3 and c = 5
Thanks for this brainstorming.