Finding extreme point of a set determined by two planes in $\mathbb R^3$

Question

Problem asks to find a extreme point the set $\{(x,y,z) \mid x-2y \leq 3 , 2y+3z \geq 4 \}$. But I don't think it has a extreme point, because it is intersection of two hyper planes in 3D, which doesn't look like has a extreme point at all. Can anyone please help me ?

Answer 1

You are right; the intersection of two closed half-spaces in $\mathbb{R}^3$ does not have any extreme points. Indeed, let $u,v$ be the normal vectors of these halfspaces. Let $w=u\times v$. The vector $w$ is parallel to the boundary of either halfspace. Hence, for any point $p$ in the intersection, the points $p\pm w$ are also there.