In a rectangular parallelepiped, a plane is drawn that passes through its diagonal, forms angles of $45^{\circ}$ and $30^{\circ}$ with the sides of the base, and is parallel to the diagonal of the base of the parallelepiped. What is the surface area of the sphere circumscribed around the parallelepiped if the distance from this plane to the diagonal of the base is $l$?
Let the plane $\alpha$ pass through the diagonal $AC_1$ of the parallelepiped $ABCDA_1B_1C_1D_1$ parallel to the diagonal $BD$ of the base $ABCD$. The plane of the base $ABCD$ passes through the line $BD$, which is parallel to the intersecting plane $\alpha$, so the line $t$ of intersection of the plane $\alpha$ with the plane of this base is parallel to the line $BD$. Let the line $t$ intersect the lines $BC$ and $CD$ at points $P$ and $Q$ respectively. Then $PB = AD$ as opposite sides of the parallelogram $ADBP$. If the lines $C_1Q$ and $DD_1$ intersect at point $N$, and the lines $C_1P$ and $BB_1$ intersect at point $M$, then the plane $\alpha$ intersects the parallelepiped along the parallelogram $AMC_1N$. From the equality of triangles $BMP$ and $B_1MC_1$ (by side and two adjacent angles), it follows that $M$ is the midpoint of the edge $BB_1$. Similarly, $N$ is the midpoint of the edge $DD_1$. Let $F$ be the foot of the perpendicular dropped from point $B$ to the line $PQ$, and $H$ be the foot of the perpendicular dropped from point $B$ to the line $MF$. Then $BH$ is the perpendicular to the plane $\alpha$. From the problem statement, it follows that $BH = l$ (since the line $BD$ is parallel to the plane $\alpha$, the distances from each point of this line to the plane $\alpha$ are equal to $l$). Moreover, $\angle HAB$ is the angle between the lateral edge $AB$ and the plane $\alpha$. According to the problem, $\angle HAB = 30^\circ$. From the right triangle $HAB$, we find that $AB = 2BH = 2l$. Reasoning similarly, we find that $AD = \frac{l}{\cos 45^\circ} = l\sqrt{2}$. Consider the right triangle $ABP$. In it,

$$
\begin{gathered}
AB = 2l, \quad PB = AD = l\sqrt{2}, \quad AP = \sqrt{4l^2 + 2l^2} = l\sqrt{6}, \\
BF = \frac{AB \cdot BP}{AP} = \frac{2l \cdot \sqrt{2}}{l\sqrt{6}} = \frac{2l}{\sqrt{3}}.
\end{gathered}
$$

Let $\angle MFB = \phi$. From the right triangle $BHF$, we find that

$$
\sin \phi = \frac{BH}{BF} = \frac{\frac{1}{2H}}{\sqrt{3}} = \frac{\sqrt{3}}{2}
$$

and since $\phi < 90^\circ$, $\phi = 60^\circ$. From the right triangle $MBF$, we find that

$$
MB = BF \tan \phi = \frac{2l}{\sqrt{3}} \cdot \sqrt{3} = 2l
$$

Then $BB_1 = 2BM = 4l$. Let $R$ be the radius of the sphere circumscribed around the parallelepiped. Since the diagonal of a rectangular parallelepiped is the diameter of this sphere, and the square of the diagonal of a rectangular parallelepiped is equal to the sum of the squares of its three dimensions,

$$
4R^2 = AB^2 + AD^2 + BB_1^2 = 4l^2 + 2l^2 + 16l^2 = 22l^2
$$

Therefore, if $S$ is the surface area of the sphere, then $S = 4\pi R^2 = 22\pi l^2$.

## Answer

$22\pi l^2$.

## Problem