= Solution
The two displacement <vectors> $\mathbf b-\mathbf a$ and $\mathbf c-\mathbf a$ lie in the required <plane>. Because the three points are non-collinear, their <cross product>
$$
\mathbf N=(\mathbf b-\mathbf a)\times(\mathbf c-\mathbf a)
$$
is nonzero and is a <normal vector> to that plane. A point with position vector $\mathbf r$ lies in the plane exactly when
$$
(\mathbf r-\mathbf a)\cdot\mathbf N=0.
$$
Expanding the <cross product> gives
$$
\mathbf N=\mathbf a\times\mathbf b+\mathbf b\times\mathbf c+\mathbf c\times\mathbf a.
$$
The terms $\mathbf a\cdot(\mathbf a\times\mathbf b)$ and $\mathbf a\cdot(\mathbf c\times\mathbf a)$ vanish, so the <equation of a plane through three points> becomes
$$
\boxed{\mathbf r\cdot
(\mathbf a\times\mathbf b+\mathbf b\times\mathbf c+\mathbf c\times\mathbf a)
=\mathbf a\cdot(\mathbf b\times\mathbf c).}
$$
Non-collinearity is important: it ensures this is a genuine plane equation with nonzero normal, rather than the vacuous equality $0=0$.
For the specified coordinates,
$$
\mathbf a\times\mathbf b=(-4,0,8),\qquad
\mathbf b\times\mathbf c=(8,0,-4),\qquad
\mathbf c\times\mathbf a=(-1,3,2).
$$
Their sum is $(3,3,6)$, with <Euclidean norm> $3\sqrt6$. Consequently either orientation of the <unit normal> is valid:
$$
\boxed{\mathbf n=\pm\frac{(1,1,2)}{\sqrt6}.}
$$
As a check, the <scalar triple product> is $12$, so the plane equation simplifies to $x+y+2z=4$, which all three supplied points satisfy. \b[The normal direction is $(1,1,2)$.]
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