= Solution
Two directions in the plane are $\mathbf b-\mathbf a$ and $\mathbf c-\mathbf a$. Their <cross product> $\mathbf N=(\mathbf b-\mathbf a)\times(\mathbf c-\mathbf a)$ is a <normal vector>. Moreover,
$$
\mathbf a\cdot\mathbf N=\mathbf a\cdot(\mathbf b\times\mathbf c)=D\ne0,
$$
so $\mathbf N$ cannot vanish. The required <unit normal> and offset are
$$
\boxed{\mathbf n=\frac{(\mathbf b-\mathbf a)\times(\mathbf c-\mathbf a)}{|(\mathbf b-\mathbf a)\times(\mathbf c-\mathbf a)|},\qquad \mu=\frac D{|(\mathbf b-\mathbf a)\times(\mathbf c-\mathbf a)|}.}
$$
Taking <dot products> with $\mathbf a,\mathbf b,\mathbf c$ gives the same offset for all three points.
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