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General method: Let $X=\{222,112\}$ be a group of two traitors we focus on. Clearly (see my answer to your earlier question) $$\mathrm{desc}(X)=\{1,2\}\times \{1,2\}\times \{2\}=\{222,122,212,112\}$$ is the set of vectors these two users can create if they collude. Since $\mathrm{desc}(X)\cap C=X$, this is not a contradiction to the code being $2-$...
Note that there are two words in $F$ which disagree on the $k^{th}$ coordinate if and only if $|X(k)|>1.$ All three words in $F$ have $0$ as their first symbol so $X(1)=\{0\}.$ Thus all descendants have 0 in their first coordinate. By definition of the direct product $\times,$ the descendants are formed by entering all elements of $X(1)$ in the first ...