= Solution
\b[The <core> is empty.] Efficiency would require $x_1+x_2+x_3=v(N)=1$. Each two-player <coalition> has value $1$, so <core> membership would also require
$$
x_1+x_2\ge1,\qquad x_1+x_3\ge1,\qquad x_2+x_3\ge1.
$$
Adding gives $2(x_1+x_2+x_3)\ge3$, contradicting efficiency, which makes the left side $2$. Thus no allocation can meet all <coalition> demands. The obstruction uses only pair <coalitions> and the grand <coalition>; imposing individual rationality cannot remove it. The equal allocation $(1/3,1/3,1/3)$ is efficient but gives each pair only $2/3<1$, so symmetry does not supply a <core> allocation.
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