Hamming distance and parity

Question

For an Hamming code of $n$ bit there are $k$ bit reserved for the data and $p$ bit for the parity where $p$ is the minimum integer for which the following inequation is satisfied: $$2^p \geqslant p + k + 1$$ This because at least $p$ bit are required to code $n$ error on a single bit $+ 1$ for the "no error" word.

For an Hamming code with distance $d$ you can detect $d-1$ bit of errors and correct $\lfloor d-1/2 \rfloor$ bit of errors.

However you can have SEC (7,4) and SEC (12,8) with the same Hamming distance $d = 3$ but the the first has $p = 3$ bit for the parity while the second has $p = 4$ bit for the parity.

The first question is what does it mean for an Hamming code H having an Hamming distance = 3 ? Does it mean that all $k$ or $n$ bit of data are required to have $d=3$ with each other ?

What is the relationship between the parity and the distance in an Hamming code ?