= Solution
The six displayed generators of the <Steane code> are independent and commuting, so the common positive eigenspace has dimension
$$
2^{7-6}=2;
$$
it encodes one logical qubit. The binary columns of the underlying Hamming parity-check matrix are all nonzero and distinct. Hence no weight-one or weight-two Pauli lies in the stabilizer normalizer outside the stabilizer. On the other hand,
$$
\overline X=X_1X_2\cdots X_7
$$
is logical, and multiplying it by $X_1X_2X_3X_4$ gives the weight-three representative
$$
\overline X\sim X_5X_6X_7.
$$
The analogous statement holds for $\overline Z$, so
$$
\boxed{d=3}.
$$
Now include both $E_1=X_5$ and $E_2=X_6X_7$ in the proposed error set. Their product is
$$
E_1^\dagger E_2=X_5X_6X_7\sim\overline X,
$$
which is non-scalar on the code and violates the <Knill--Laflamme condition>. Since multiplying by a logical operator does not change commutation with stabilizers, $X_5$ and $X_6X_7$ have the same <error syndrome>. A decoder cannot know which occurred and may apply a correction that leaves a logical $\overline X$ error.
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