 Solution

With normal vector $n=(-1,-2,2)$, a plane through $A=(2,-1,1)$ has equation $n\cdot (r-A)=0$. As $n\cdot A=2$, its plane equation isThe distance from a point to a plane is the absolute signed offset divided by the Euclidean norm of its normal vector. Hence the Euclidean distance from the origin is $\boxed{2/3}$, since $|n|=3$. The closest point is $2n/9=(-2,-4,4)/9$, whose Euclidean distance from the origin is $2/3$ and which satisfies the plane equation.