= Solution
One choice of the <Hamming code of length seven> is the <linear map>
$$
h(a,b,c,d)=(a+b+d,\ a+c+d,\ a,\ b+c+d,\ b,\ c,\ d)
$$
over $\mathbb F_2$. Its image is the <kernel> of the <parity-check matrix>
$$
H=\begin{pmatrix}1&0&1&0&1&0&1\\0&1&1&0&0&1&1\\0&0&0&1&1&1&1\end{pmatrix}.
$$
Indeed $Hh(a,b,c,d)^T=0$, the encoder is <injective> because positions $3,5,6,7$ recover its inputs, and $H$ has <rank> three. Its columns are nonzero and pairwise distinct. Hence no nonzero <codeword> has <Hamming weight> one or two. The word $h(1,0,0,0)=(1,1,1,0,0,0,0)$ has weight three, proving that the <minimum Hamming distance> is \b[three].
If $r=h(a,b,c,d)+e_j$ is received with one erroneous bit, its <syndrome> $Hr^T$ is column $j$ of $H$. These seven columns are distinct, so the <syndrome> identifies the bit to flip. A zero <syndrome> indicates no error under the assumption of at most one error.
For the <extended Hamming code>, append $\sum_{j=1}^7h_j=a+b+c$. Every extended <codeword> has <even> weight. Since every original nonzero word has weight at least three, its extended weight is at least four; the weight-three example attains four. Thus
$$
\boxed{d_{\min}=4.}
$$
The <minimum-distance error-detection and correction guarantee> gives \b[detection of up to three errors and correction of one error]. Detection here means rejecting a word outside the code, rather than simultaneously identifying and correcting every detected pattern; the usual combined decoder corrects single errors and flags double errors.
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