= Solution
Identification is stronger than <quantum state exclusion>. Here an outcome only needs to rule out its associated preparation. Use the following <orthonormal basis> for a <basic measurement>
$$
|b_0\rangle=\frac{|0\rangle-|1\rangle}{\sqrt2},\qquad
|b_1\rangle=\frac{|0\rangle+|1\rangle}{\sqrt2},\qquad
|b_2\rangle=|2\rangle.
$$
Its defining overlaps satisfy $\langle b_m|\phi_m\rangle=0$ for $m=0,1,2$. By the <Born rule>, observing $b_m$ makes preparation $\phi_m$ impossible. \b[Measure in this basis and send its outcome label $m$; the prize is guaranteed.]
For completeness, the <conditional probability> matrix, with rows $m$ and columns the prepared-state label $j$, is
$$
\bigl[\Pr(m\mid\phi_j)\bigr]_{m,j}
=\begin{pmatrix}0&1/2&1/2\\1/2&0&1/2\\1/2&1/2&0\end{pmatrix}.
$$
Every column sums to one and the diagonal vanishes. Each result leaves two possible preparations, so this <projective state exclusion> does not contradict the impossibility of perfect identification. The family consists of <antidistinguishable quantum states>.
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