For the second term
C(n,1) (- a) = -12
n! / (n-1)! * a = 12 { n! /(n -1)! = (n) }
n * a = 12
a = 12/n → a^2 = 144/n^2
For the third term
C (n,2) (-a)^2 = 54
n! / [ (n -2)! (2!) ] * a^2 = 54 { n! / (n -2)! = (n)(n -1) }
(n)(n-1) * 144/n^2 = 108
(n^2 - n) / n^2 = 108 / 144 = 3 / 4
1n^2 - n = (3/4)n^2
(1/4)n^2 - n = 0
n (1/4 * n - 1) = 0
n = 0 reject
n = 4 accept
Since the expansion terms alternate signs, then
a = -12/n = -12/4 = -3
The binomial is (1 -3x)^4
c = C(4, 3) * (-3)^3 = 4 * -27 = -108