= Solution
Expanding the squared <Euclidean distances> gives
$$
|\mathbf x-\mathbf a|^2=|\mathbf x-\mathbf b|^2
\iff
2(\mathbf b-\mathbf a)\mathbin\cdot\mathbf x
=|\mathbf b|^2-|\mathbf a|^2.
$$
Thus one may take
$$
\boxed{\mathbf n_{AB}=\frac{\mathbf b-\mathbf a}{|\mathbf b-\mathbf a|},
\qquad
p_{AB}=\frac{|\mathbf b|^2-|\mathbf a|^2}{2|\mathbf b-\mathbf a|}}.
$$
The normal is parallel to $\overrightarrow{AB}$, so $L_{AB}$ is its <perpendicular bisector>. If $\mathbf x$ lies on both $L_{AB}$ and $L_{BC}$, then $|\mathbf x-\mathbf a|=|\mathbf x-\mathbf b|=|\mathbf x-\mathbf c|$, hence it also lies on $L_{CA}$. Geometrically, the three perpendicular bisectors of the noncollinear triangle concur at its <circumcenter>.
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