Assume that $X,Y,Z$ are three hypercuboid with the same dimension $k$. Specifically, we have $X=\{x_1,x_2,...,x_k:0\leq x_i\leq c_i, \forall\,i\}$ and $Y=\{y_1,y_2,...,y_k:0\leq y_i\leq b_i, \forall\,i\}$ and $Z=\{z_1,z_2,...,z_k:0\leq z_i\leq a_i, \forall\,i\}$ where each $c_i$ and $b_i$ and $a_i$ are constant positive real number. My question is whether the following equality holds or not: $$\text{Conv}(\text{extr}(X\cup Y\cup Z))=\text{Conv}(\text{extr}(X)\cup\text{extr}(Y)\cup\text{extr}(Z))$$ where $\text{Conv}$ means convex hull and $\text{extr}(X)$ means the extreme points of the set $X$.
I conjecture that the equality is also true by drawing a toy example with $k=2$. However, I cannot argue it mathematically. Is there any hint or steps? Thank you!
Edit: The definition of the extreme point is that a point in a compact set $A$ is an extreme point if this point cannot be any linear combination of any other two distinct points in $A$.