[proofplan] We compute the syndrome of a vector from its support: for a binary vector, the parity-check value is the sum of the columns indexed by the nonzero coordinates. Thus a weight-$3$ vector is a Hamming codeword exactly when the three corresponding nonzero vectors in $\mathbb{F}_2^3$ sum to $0$. Over $\mathbb{F}_2$, three distinct nonzero vectors sum to $0$ exactly when they have the form $\{u, v, u+v\}$, which is precisely a Fano line. [/proofplan]

[step:Express the syndrome of a weight-$3$ vector as the sum over its support] Let \begin{align*} P := \mathbb{F}_2^3 \setminus \{0\} \end{align*} denote the set of coordinate labels. For a vector $c = (c_p)_{p \in P} \in \mathbb{F}_2^P$, define its support by \begin{align*} \operatorname{supp}(c) := \{p \in P : c_p = 1\}. \end{align*} Since the coordinates are binary, the Hamming weight of $c$ is $|\operatorname{supp}(c)|$. By the definition of $H$, \begin{align*} H(c) = \sum_{p \in P} c_p p. \end{align*} Because $c_p = 0$ for $p \notin \operatorname{supp}(c)$ and $c_p = 1$ for $p \in \operatorname{supp}(c)$, this becomes \begin{align*} H(c) = \sum_{p \in \operatorname{supp}(c)} p. \end{align*} Therefore, if $c$ has Hamming weight $3$ and \begin{align*} \operatorname{supp}(c) = \{u, v, w\} \end{align*} with $u, v, w \in P$ distinct, then \begin{align*} c \in \operatorname{Ham}(7,2) \iff H(c)=0 \iff u+v+w=0. \end{align*} [/step]

[step:Characterize the zero-sum triples in $\mathbb{F}_2^3$]Let $u, v, w \in P$ be distinct. Then \begin{align*} u+v+w=0 \end{align*} if and only if \begin{align*} w = u+v. \end{align*} Indeed, adding $u+v$ to both sides of $u+v+w=0$ gives \begin{align*} w = u+v, \end{align*} because every element of $\mathbb{F}_2^3$ is its own additive inverse. Conversely, if $w=u+v$, then \begin{align*} u+v+w = u+v+(u+v)=0. \end{align*} The vector $u+v$ is nonzero because $u \neq v$, and it is distinct from both $u$ and $v$: if $u+v=u$, then $v=0$, and if $u+v=v$, then $u=0$, contradicting $u,v \in P$.[/step]

[guided]We need to understand exactly which triples of nonzero column labels can have zero syndrome. Let $u, v, w \in P$ be distinct vectors. The equation \begin{align*} u+v+w=0 \end{align*} can be solved for $w$ by adding $u+v$ to both sides. In $\mathbb{F}_2^3$, addition is componentwise modulo $2$, so each vector is its own additive inverse. Hence \begin{align*} w = u+v. \end{align*} Conversely, suppose $w=u+v$. Then substituting this expression into the sum gives \begin{align*} u+v+w = u+v+(u+v)=0, \end{align*} again because each term appears twice and $a+a=0$ in characteristic $2$. We also verify that this really gives a triple of Fano-plane points. Since $u$ and $v$ are distinct, $u+v \neq 0$; otherwise $u=v$. Thus $u+v \in P$. Moreover $u+v$ cannot equal $u$, because then $v=0$, and it cannot equal $v$, because then $u=0$. Both are impossible since $u,v \in P$. Therefore the zero-sum triples are exactly the triples $\{u,v,u+v\}$ with $u,v \in P$ distinct.[/guided]

[step:Identify the zero-sum triples with the Fano lines] In the Fano plane identified with the nonzero vectors of $\mathbb{F}_2^3$, the line through two distinct points $u,v \in P$ is the set of nonzero vectors in the two-dimensional subspace $\operatorname{span}_{\mathbb{F}_2}\{u,v\}$. Since the field is $\mathbb{F}_2$, this subspace has exactly the four vectors \begin{align*} 0,\quad u,\quad v,\quad u+v. \end{align*} Therefore the corresponding projective line is \begin{align*} L(u,v) := \{u,v,u+v\}. \end{align*} Thus the supports of weight-$3$ codewords in $\operatorname{Ham}(7,2)$ are exactly the Fano lines. [/step]

[step:Verify both inclusions for incidence vectors] First let $c \in \operatorname{Ham}(7,2)$ have Hamming weight $3$. By the first step, if \begin{align*} \operatorname{supp}(c)=\{u,v,w\}, \end{align*} then $u+v+w=0$. By the zero-sum characterization, after relabelling if necessary, \begin{align*} w=u+v. \end{align*} Hence \begin{align*} \operatorname{supp}(c)=\{u,v,u+v\}=L(u,v), \end{align*} so $c=\mathbb{1}_{L(u,v)}$ is the incidence vector of a Fano line. Conversely, let $L=L(u,v)=\{u,v,u+v\}$ be a Fano line with $u,v \in P$ distinct. Its incidence vector $\mathbb{1}_L \in \mathbb{F}_2^P$ has Hamming weight $3$, since the three points $u$, $v$, and $u+v$ are distinct. Its syndrome is \begin{align*} H(\mathbb{1}_L) = \sum_{p \in L} p = u+v+(u+v)=0. \end{align*} Therefore $\mathbb{1}_L \in \ker H = \operatorname{Ham}(7,2)$. Finally, there are exactly seven such lines. There are $\binom{7}{2}=21$ unordered pairs of distinct points in $P$, and each line $L(u,v)=\{u,v,u+v\}$ contains exactly three unordered pairs of its points. Since any two distinct points determine the unique line $L(u,v)$, the number of distinct Fano lines is \begin{align*} \frac{\binom{7}{2}}{3}=7. \end{align*} This proves that the weight-$3$ codewords of $\operatorname{Ham}(7,2)$ are exactly the incidence vectors of the seven Fano lines. [/step]