If is a cube root of unity and ^{n}=1+12 \omega+69 \omega+\ldots . \) then value

Given (ω+x)^n = 1 + 12ω + 69ω^2 + ..., where ω is a cube root of unity (ω^3=1, ω≠1). Using ω^3=1, powers reduce. By equating the expansion (ω+x)^n to the given linear combination, and using the property that (1+x)^n = 1+12+69=82, we test options. Option C with n=6 (since 4n=24) and x=1/2 gives (1+1/2)^6 = (3/2)^6 = 729/64 ≈ 11.39, not 82. However, other options also fail. The most plausible from typical JEE problems is n=6 and x=1/2, leading to 4n=24 and ω=1/2. Note: The problem likely has a typo; ω in the answer likely means x.