Answer: 510

Small Hint:

The first inequality rearranges to $(x - y)(1 + z) \lt 0,$ and the second to $(y - z)(1 + x) \lt 0$

Big Hint:

The bounded piece is $-1 \lt x \lt y \lt z$ on the plane; its corners come from the boundaries $x = -1,$ $x = y,$ and $y = z$

Solution:

Since $x - yz - (y - zx)$ $= (x - y) + z(x - y)$ $= (x - y)(1 + z),$ and similarly $y - zx - (z - xy)$ $= (y - z)(1 + x),$ the conditions are $$(x - y)(1 + z) \lt 0$$ and $$(y - z)(1 + x) \lt 0.$$ Each condition offers two sign patterns, giving four combinations. The combination $x \gt y,$ $z \lt -1,$ $y \gt z,$ $x \lt -1$ is impossible on the plane: $x, z \lt -1$ forces $y \gt 77,$ contradicting $x \gt y.$ Two of the remaining combinations allow a coordinate to run off to infinity, producing the two unbounded regions.

The bounded region is $x \lt y,$ $y \lt z,$ $x \gt -1$ (the fourth constraint $z \gt -1$ is then automatic): the set $-1 \lt x \lt y \lt z$ on the plane. Its closure is the triangle whose vertices come from intersecting the boundary lines pairwise: $x = -1,$ $x = y$ gives $(-1, -1, 77);$ $x = -1,$ $y = z$ gives $(-1, 38, 38);$ and $x = y = z$ gives $(25, 25, 25).$

With $A = (-1, -1, 77),$ the edge vectors are $B - A = (0, 39, -39)$ and $C - A = (26, 26, -52),$ whose cross product is $-1014\,(1, 1, 1),$ of length $1014\sqrt{3}.$ The area is $\frac{1014\sqrt{3}}{2} = 507\sqrt{3},$ so $a + b = 507 + 3 = 510.$