= Solution
The PDF places the extra $X$ on $q_2$ after that qubit has interacted with $a_1$ but before it interacts with $a_2$. Its timing is essential. With no incoming error, the first extracted parity is zero. The fault then changes the data to
$$
u|010\rangle+v|101\rangle.
$$
The remaining parity measurement gives $a_2=1$ on both branches. Thus the measured <error syndrome> is $01$, which the rule in the preceding part interprets as an error on $q_1$. Applying $X_1$ leaves
$$
\boxed{u|110\rangle+v|001\rangle=X_1X_2|\psi_L\rangle.}
$$
This is outside the original encoded subspace: neither of its supporting basis strings is $000$ or $111$. The incorrect syndrome causes a second data error rather than restoring the input. This <timed bit flip during repetition-code syndrome extraction> demonstrates why a simple parity circuit need not tolerate its own faults.
The same failure can be checked for every incoming channel error. Let $f\in\{0,1\}$ indicate the internal fault and let $c(s)$ be the correction bit pattern from the table. The observed syndrome and the final pre-correction data pattern are
$$
s_{\mathrm{obs}}=s(e)\oplus(0,f),\qquad d=e\oplus(0,f,0).
$$
Because $s$ is linear over binary addition and $s(c(s_{\mathrm{obs}}))=s_{\mathrm{obs}}$, the residual pattern $r=d\oplus c(s_{\mathrm{obs}})$ satisfies
$$
\boxed{s(r)=s(d)\oplus s_{\mathrm{obs}}=(f,0).}
$$
Whenever the fault occurs, the residual syndrome is $10$, so recovery never returns the state to the encoded subspace, even if a channel error also occurred. The diagram therefore must not be treated as an extra independent pre-extraction data flip that would always be diagnosed normally.
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