= Solution
Number the three data <qubits> from top to bottom. It suffices initially to follow a pure input $|\psi\rangle=a|0\rangle+b|1\rangle$; linearity will extend the result to every <density operator>. The first two <controlled-NOT gates> create $a|000\rangle+b|111\rangle$. The first layer of <Hadamard gates> then encodes this as
$$
a|+++\rangle+b|---\rangle,
$$
the three-<qubit> <phase-flip repetition code>. Let $e_i$ indicate whether a <Pauli Z gate> occurred on data <qubit> $i$ during storage. The next layer of <Hadamard gates> uses $HZH=X$ and gives
$$
a|e_1e_2e_3\rangle+b|1\oplus e_1,1\oplus e_2,1\oplus e_3\rangle.
$$
The two initially zero syndrome <quantum ancillas> receive the adjacent parities. In the diagram's top-to-bottom order, their <error syndrome> is
$$
s_1=e_1\oplus e_2,\qquad s_2=e_2\oplus e_3.
$$
Both terms of the logical superposition have these same parities, so the syndrome is independent of $a,b$. Measuring it reveals no logical-state information.
At the marked position of $R$ in the printed circuit, there has already been another layer of <Hadamard gates> on the data. There is also a layer immediately after $R$. Thus a desired correction $X_i$ in the parity-extraction frame must be implemented as $R=Z_i$ at the actual recovery location, since $H^{\otimes3}Z_iH^{\otimes3}=X_i$. Choose the recovery according to
$$
\boxed{\begin{array}{c|c|c|c}
(s_1,s_2)&\text{weight-zero or one pattern}&\text{complementary pattern}&R\\\hline
00&000&111&I\\
10&100&011&Z_1\\
11&010&101&Z_2\\
01&001&110&Z_3
\end{array}}
$$
The two patterns on each row share a syndrome because complementing all three bits does not change either parity. For a weight-zero or one error, the recovery removes the entire error in the parity-extraction frame. The remaining state is $a|000\rangle+b|111\rangle$, and the two final <controlled-NOT gates> give $|\psi\rangle|00\rangle$.
For a weight-two or three error, minimum-weight recovery instead leaves all three bits flipped, giving $a|111\rangle+b|000\rangle$. The final <controlled-NOT gates> send this to $(a|1\rangle+b|0\rangle)|00\rangle=X|\psi\rangle|00\rangle$. Consequently, for every input <density operator>,
$$
\boxed{\rho'=\rho\quad(k=0,1),\qquad\rho'=X\rho X\quad(k=2,3).}
$$
The logical failure is a <Pauli X gate> after decoding, even though the stored physical errors were <Pauli Z gates>: $Z_1Z_2Z_3$ exchanges the two encoded codewords. This is the <logical failure of the three-qubit phase-flip repetition code>.
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