Let $X$ be a point contained in all three squares. The distance from $X$ to any point in any of the three squares is at most $\sqrt{2}$, the length of the diagonal of the squares. Therefore, triangle $A B C$ is contained in a circle of radius $\sqrt{2}$, so its circumradius is at most $\sqrt{2}$. The triangle with greatest area that satisfies this property is the equilateral triangle in a circle of radius $\sqrt{2}$. (This can be proved, for example, by considering that the maximum altitude to any given side is obtained by putting the opposite vertex at the midpoint of its arc, and it follows that all the vertices are equidistant.) The equilateral triangle is also attainable, since making $X$ the circumcenter and positioning the squares such that $A X, B X$, and $C X$ are diagonals (of the three squares) and $A B C$ is equilateral, leads to such a triangle. This triangle has area $3 \sqrt{3} / 2$, which may be calculated, for example, using the sine formula for area applied to $A B X, A C X$, and $B C X$, to get $3 / 2(\sqrt{2})^{2} \sin 120^{\circ}$.