 Solution

Only $z,w$ have printed nonnegativity restrictions; $x,y$ are unrestricted. In four-dimensional space, a vertex of a linear polyhedron needs four independent active constraints. There are exactly four available constraints, so all must be active. Solving them givesand therefore the only possible vertex of a linear polyhedron is $(x,y,z,w)=(2,3,0,0)$. It is feasible, and its four active normals are independent, so it is indeed a vertex of a linear polyhedron.

One can also see uniqueness geometrically from the slack variables $s_1=1-2x+y-5w$ and $s_2=3-y-z-w$. The feasible region is in affine bijection with the nonnegative orthant in $(s_1,s_2,z,w)$:That orthant has only the origin as an extreme point, giving the same unique vertex of a linear polyhedron.