If the sum of the coefficients in the expansion of (x−2y+3z)n,n∈ℕ is 128 , then the  greatest coefficient in the expansion of (1+x)n is

Sum of the coefficients in the expansion(x−2y+3z)n is (1−2+3)n=2n (∵ Put x=y=z=1) i.e. 2n=128⇒n=7 therefore, greatest coefficient in the expansion of (1+x)7 is 7C3 or 7C4⇒7C3=7C4=35