To solve this system of equations, we can use the method of substitution.

From the first equation, we can solve for a in terms of b:
a = 14 - b

Substituting this into the third equation, we have:
(14 - b) + c = 11
c = 11 - 14 + b
c = b - 3

Now we can substitute these expressions for a and c into the second equation:
b + (b - 3) + 15 = 0
2b + 12 = 0
2b = -12
b = -6

Substituting this value of b into the first equation, we can solve for a:
a + (-6) = 14
a = 14 + 6
a = 20

Finally, substituting the values of a and b into the expression for c, we can solve for c:
c = (-6) - 3
c = -9

Therefore, the solution to the system of equations is a = 20, b = -6, and c = -9.