$\text{ Plane }\mathrm{x}+\mathrm{y}+\mathrm{z}=\text{ a cuts the unit cube with one vertex at origin and three adjacent vertices at positive x, y, z axes.}$

$\text{ for }\mathrm{a}\in (1,2)\text{ the area of intersection is a hexagon with angles }{\mathrm{\theta }}_1,{\mathrm{\theta }}_2,\dots \dots ,{\mathrm{\theta }}_6\text{ such that }\cos {\mathrm{\theta }}_1+\cos {\mathrm{\theta }}_2\dots .+\cos {\mathrm{\theta }}_6=-3.$

$\text{ For }\mathrm{a}=\frac{3}{2}\text{ area enclosed by the intersection is }\frac{3\sqrt{3}}{4}$

$\text{ For }\mathrm{a}=\frac{4}{3}\text{ area enclosed by the intersection is }\frac{13\sqrt{3}}{18}$

$\text{ Maximum area of intersection is when }\mathrm{a}=1\text{ or }2\text{. }$