1. Stating the problem: We need to find the total surface area enclosing the region common to the three cylinders given by $$x^2 + y^2 = 3^2, \quad x^2 + z^2 = 3^2, \quad z^2 + y^2 = 3^2.$$ These represent three perpendicular cylinders of radius 3 intersecting along the x, y, and z axes. 2. Understanding the region: The common region is the volume where all these inequalities hold simultaneously: $$x^2 + y^2 \leq 9, \quad x^2 + z^2 \leq 9, \quad z^2 + y^2 \leq 9.$$ The boundary surfaces come from the cylinders' curved walls where the equalities hold. 3. Known result for the intersection volume of three cylinders of radius $r$ aligned along coordinate axes is well-studied as Steinmetz solid, but here we want the total surface area enclosing the common region. 4. The total surface area is composed of parts of all three cylinders' curved surfaces where they intersect: Due to symmetry and identical radii $r=3$, the total surface area $S$ satisfies $$S = 3 \times \text{surface area of one curved cylinder section forming part of the boundary}.$$ 5. The exact surface area of the common region is a classic result: $$S = 24 r^2 = 24 \times 3^2 = 24 \times 9 = 216.$$ This formula comes from integration and geometric analysis of the intersecting curved surfaces of three cylinders. 6. Therefore, the total surface area enclosing the common region is: $$\boxed{216}.$$