Use normals for parallel planes and lines inside a plane

$(-2,1,1)\times(1,-3,2)=(5,5,5)=5(1,1,1)$. Thus Π:$\mathbf r\cdot(1,1,1)=(2,0,0)\cdot(1,1,1)=2$. For B,$(6,-5,1)\cdot(1,1,1)=6-5+1=2$, verifying membership.

The normal through C is $\mathbf r=(-2,1,-3)+\lambda(1,1,1)$. Intersection with Π gives $(-2+\lambda)+(1+\lambda)+(-3+\lambda)=2$, so $3\lambda-4=2$, λ=2 and $ON=(0,3,-1)$.

Choose N′ with C the midpoint of NN′. Then $ON′=2OC-ON=2(-2,1,-3)-(0,3,-1)=(-4,-1,-5)$. A plane parallel to Π through N′ is Π1:$\mathbf r\cdot(1,1,1)=-10$. C has equal distance $6/\sqrt3$ to the two distinct planes.

$(\mathbf u+\mathbf v)\cdot(\mathbf u\times\mathbf v)=\mathbf u\cdot(\mathbf u\times\mathbf v)+\mathbf v\cdot(\mathbf u\times\mathbf v)=0$. Each term vanishes because the cross product is perpendicular to both factors. Therefore l’s direction lies parallel to P; both contain the origin, so l lies in P. Nonparallel u,v ensure a nonzero plane normal and a nonzero line direction.

$\mathbf u+\mathbf v=(2,-1,3)$ and $\mathbf u\times\mathbf v=(7,5,-3)$. Their cross product is $(2,-1,3)\times(7,5,-3)=(-12,27,17)$, perpendicular to l and lying in P. Thus the required line is $\mathbf r=\mu(-12,27,17)$, μ real.