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The sum of the surface areas of the polyhedra into which a parallelepiped is divided by sections equals the sum of the surface area of the parallelepiped and the areas of the internal surfaces. The sum of the areas of the internal surfaces is twice the sum of the areas of the sections.

Find the maximum possible area of a section passing through the diagonal \(XY\) of an arbitrary parallelepiped with edges \(a \leq b \leq c\). The section is a parallelogram \(ZXTY\), whose vertices lie on the opposite edges of the parallelepiped. The area of the parallelogram is equal to the product of the length of the diagonal \(XY\) by the distance from point \(Z\) to \(XY\).

Therefore, the section passes through one of the parallelepiped's edges. Thus, the largest area is one of the diagonal sections, and all of these sections are rectangles. Find the largest of their areas:
\[ S_{1}=a \sqrt{b^{2}+c^{2}}, \quad S_{2}=b \sqrt{a^{2}+c^{2}}, \quad S_{3}=c \sqrt{b^{2}+a^{2}} \]

Given \( a \leq b \leq c \), it follows that \(a^{2} b^{2}+a^{2} c^{2} \leq c^{2} b^{2}+a^{2} c^{2}\) and \(a^{2} b^{2}+c^{2} b^{2} \leq c^{2} b^{2}+a^{2} c^{2}\). Therefore, \( S_{1} \leq S_{3} \) and \( S_{2} \leq S_{3} \). Hence, the maximum area is of the section passing through the largest edge. The sum of the surface areas of the polyhedra into which the parallelepiped is divided by these sections is:
\[ 2\left(A A_{1} \cdot A B+A A_{1} \cdot A D+A B \cdot A D\right)+4 \cdot 150=1116 \]

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