 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 ofis 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:It is orthogonal to each normal vector. Since both planes contain $A$, the whole intersection line is $r=(2,-1,1)+t(6,2,5)$, $t\in \mathbb{R}$.