What point of the plane $3x+2y+z-6=0$ in $3$-dimensional space is closest to the origin when the distance is measured by each of the following three norms: the $1$-norm, the $2$-norm, the $\infty$-norm?
For $2$-norm, I know that we can calculate the minimum distance between the origin and the plane by formula and then set the line that is cross the origin and perpendicular to the plane, then we can get the point we want. But when it comes to 1-norm and $\infty$-norm, it does not work.
For the $1$-norm the surfaces of equal distance from the origin (the equivalent of spheres) are octahedrons with axes aligned with the co-ordinate axes. As the "radius" increases from $0$, the first surface to intersect a given plane will intersect it at a vertex (and possibly simultaneously at more than one vertex for some planes). So the minimum distance will lie along one of the lines $x=y=0$, $x=z=0$ or $y=z=0$.
For the $\infty$-norm the surfaces of equal distance from the origin are cubes with faces aligned with the co-ordinate axes. Again, as the "radius" increases from $0$, the first surface to intersect a given plane will intersect it at a vertex of the cube. So the minimum distance will lie along the one of the lines $x=y=z$, $x=-y=z$ or $x=y=-z$.