Surface area of a tricylinder ($A=24a^2(2-\sqrt{2})$)

For three radius-$a$ circular cylinders with mutually perpendicular axes through one point, the boundary of their common intersection has three congruent cylindrical surface pieces. On $x^2+y^2=a^2$, the height restriction is $|z|\le a\min (|\sin \theta |,|\cos \theta |)$. Eight congruent sectors give surface area $8a^2(2-\sqrt{2})$ per piece. Pairwise intersections are curves of zero surface area, so the three pieces add.