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Find the maximum possible area of a cross-section passing through the diagonal \( XY \) of an arbitrary parallelepiped with edges \( a \leq b \leq c \).

The cross-section is a parallelogram \( ZXTY \), where the vertices lie on opposite edges of the parallelepiped. The area of the parallelogram is the product of the length of the diagonal \( XY \) and the distance from point \( Z \) to the line \( XY \).

By projecting the parallelepiped onto a plane perpendicular to the diagonal \( XY \), the largest distance from any point \( Z \) to the line \( XY \) occurs when \( Z \) coincides with one of the vertices \( A, B, \) or \( C \).

Thus, the cross-section with the largest area passes through one of the parallelepiped’s edges, forming a rectangle. The maximal areas of these sections are:

\[ S_1 = a \sqrt{b^2 + c^2}, \quad S_2 = b \sqrt{a^2 + c^2}, \quad S_3 = c \sqrt{a^2 + b^2}. \]

Given \( a \leq b \leq c \), it follows that:
\[ S_1 \leq S_3 \quad \text{and} \quad S_2 \leq S_3. \]

Thus, the largest area is \( S_3 \), which corresponds to the section passing through the longest edge. The area is calculated as:

\[ 20 \sqrt{5^2 + 12^2} = 260. \]

Therefore, the maximum area of the cross-section is \( 260 \).

The sum of the surface areas of the polyhedra, into which the parallelepiped is divided by these sections, is:

\[ 2(AA_1 \times AB + AA_1 \times AD + AB \times AD) + 4 \times 260 = 1840. \]

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