 Solution

For a coalition $S$, its coalition excess at an allocation is $e(S,x)=v(S)-x(S)$, measuring its complaint. Sort all these excesses of a coalition in decreasing order. The nucleolus is the imputation whose sorted vector is smallest in lexicographic order: minimize the greatest complaint first, then the next among ties, and continue. Including the empty and grand coalitions merely inserts two fixed zero entries and does not change the minimizing allocation.

For every efficient allocation, the excesses of a coalition of $\{2\}$ and $\{1,3\}$ sum to zero:The largest proper-coalition coalition excess is therefore at least zero. Since the core of a cooperative game from part (b) is nonempty, this first-stage minimum is exactly zero and its minimizers are precisely that core of a cooperative game. Restrict to $x=(t,2,10-t)$, $1\le t\le 6$. The six proper-coalition excesses of a coalition areThe two zero excesses of a coalition are fixed. Among the remaining excesses of a coalition, $t-7<t-6$, so the next largest complaint is $\max \{1-t,t-6\}$. It is minimized by equalizing those two affine expressions:This minimizer is unique, so later lexicographic stages cannot change it. ThusIts sorted proper-coalition coalition excess vector is $(0,0,-5/2,-5/2,-5/2,-7/2)$. The calculation both identifies the nucleolus and proves uniqueness for this game without replacing the requested computation by the general uniqueness theorem.