 Solution

Take the parity-check matrixover ${\mathbb{F}}_2$. Its columns are all the nonzero vectors of ${\mathbb{F}}_2^3$. The Hamming code is $C=\ker H$. Since $H$ has rank three, $C$ is a binary linear code of length seven and dimension four. No nonzero codeword has weight one or two, since no column is zero and no two columns coincide. Columns $u,v,u+v$ give a codeword of weight three, so $C$ has parameters $[7,4,3]$.

For a received vector $w=c+e$, its syndrome is $Hw=He$. A single error in position $j$ gives exactly column $j$ of $H$, so the syndrome identifies that position uniquely; zero syndrome means no error under the one-error assumption. The radius-one Hamming balls around the $16$ codewords are disjoint and have total size $16(1+7)=128=2^7$, so this is also a perfect code.

There are $\left(\genfrac{}{}{0ex}{}{7}{2}\right)/3=7$ weight-three codewords: every pair of columns determines the third, and each triple is counted three times. The all-ones vector is a codeword because the column sum is zero, so complementation gives seven weight-four codewords. Together with zero and the all-ones vector these account for all sixteen. The weight enumerator is therefore