= Solution
For an <imputation> $y$, the <excess of a coalition> is $e(S,y)=v(S)-\sum_{i\in S}y_i$. Arrange these excesses in decreasing order. The <nucleolus> is the <imputation> that lexicographically minimizes this ordered vector: minimize the largest complaint first, then the next largest among ties, and so on. The empty and grand-coalition excesses are fixed at zero for every efficient allocation, so they can be omitted without changing the lexicographic comparison.
For every efficient $y$ the three pair excesses have sum
$$
e(\{1,2\},y)+e(\{1,3\},y)+e(\{2,3\},y)=6+9+11-2(15)=-4.
$$
Hence at least one pair has excess at least $-4/3$, and the largest proper-coalition excess is at least $-4/3$. At the proposed allocation all three pair excesses equal $-4/3$, while all three singleton excesses equal $-8/3$. It therefore attains the best possible largest proper-coalition excess.
Moreover, any allocation attaining this bound must have all three pair excesses equal $-4/3$, since they are individually at most that value and sum to $-4$. This forces
$$
y_1+y_2=\frac{22}3,\qquad y_1+y_3=\frac{31}3,\qquad y_2+y_3=\frac{37}3,
$$
whose unique solution is $y=x$. It is already an <imputation>, so the first nonconstant lexicographic stage has a unique minimizer; there are no later choices to resolve. Therefore
$$
\boxed{\operatorname{nuc}(v)=x:\quad\text{(c) is true.}}
$$
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