= Solution
The displacement <vectors> are $B-A=(-1,2,-2)$ and $C-A=(1,0,1)$. Their <cross product> is $(2,-1,-2)$, with <Euclidean norm> three. Thus either sign of
$$
\boxed{\hat n_2=\frac{(2,-1,-2)}3}
$$
is a <unit normal vector> to the second <plane>, whose equation is $2x-y-2z=3$. A <direction vector> for the intersection is the <cross product> of the two <normal vectors>:
$$
\boxed{(-1,-2,2)\times(2,-1,-2)=(6,2,5).}
$$
It is <orthogonal> to each <normal vector>. Since both <planes> contain $A$, the whole intersection line is $r=(2,-1,1)+t(6,2,5)$, $t\in\mathbb R$.
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