= Solution
The columns of the <parity-check matrix> of the binary <Hamming code> are precisely the $n=2^d-1$ distinct nonzero vectors of $\mathbb F_2^d$. If $h_1\ne h_2$, then $h_1+h_2$ is nonzero and differs from both summands, so it occurs as a unique third column $h_3$.
A word supported on three positions is a codeword exactly when the corresponding three columns sum to zero. Every unordered pair of columns determines one such triple, and each triple is counted once by each of its three pairs. Therefore
$$
\boxed{\nu(3)=\frac{1}{3}\binom n2=\frac{n(n-1)}6}.
$$
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