## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-257.jpg?height=464&width=514&top_left_y=441&top_left_x=446)

Let there be a cube $T_{1}=A B C D K L M N$ and a cube $T_{2}$, obtained by rotating $T_{1}$ around $E F$ by $90^{\circ}$. Let $P_{1}$ and $P_{2}$ be points on the edges $ND$ and $N K$, at a distance of $a / \sqrt{2}$ from $N$, and $P_{3}$ and $P_{4}$ be points on the edges $M L$ and $M C$, at a distance of $a / \sqrt{2}$ from $M$. Then $P_{1} P_{2} P_{3} P_{4}$ is a square, since $P_{1} P_{2}=P_{3} P_{4}=a$.

Moreover, $A N \perp P_{1} P_{2}, A N \perp P_{1} P_{4} \Rightarrow A N \perp P_{1} P_{2} P_{3} P_{4} \quad$ (Theorem 11) $\Rightarrow$ $\Rightarrow E F \perp P_{1} P_{2} P_{3} P_{4} \Rightarrow E F \perp R P_{i}, i=1,2,3,4$, if $R$ is the intersection point of $E F$ with $R_{1} P_{2} P_{3} P_{4}$. Then $R$ is the center of the square due to the equality of the inclined lines $F P_{1}, i=1,2,3,4 \Rightarrow \angle P_{2} R P_{1}=90^{\circ}$. This means that when $T_{1}$ is rotated by $90^{\circ}$ around $E F$ clockwise (if viewed from $E$ to $F$), the point $P_{i}$ transitions to $P_{i+1}, i=1,2,3, P_{4}$ transitions to $P_{1}$. Similarly, if the points $Q_{i}$ are symmetric to $P_{i}$ relative to the plane $L C D K$, then upon rotating by $90^{\circ}$ around $E F$, $Q_{i}$ transitions to $Q_{i+1}, i=1,2,3, Q_{4}$ transitions to $Q_{1}$.

Now we can trace how the faces of the cube $T_{2}$ will pass, considering the immobility of points $E$ and $F$. Thus, since a plane is uniquely determined by its three points not lying on the same line, we conclude: the face $BLMC$ after rotation will pass through the points $P_{1}, P_{4}, Q_{1}, Q_{4}$; the face $C M N D$ - through the points $P_{1}, P_{2}, F$. The behavior of the other faces is similar. In the end, we have: the faces of the cube $T_{2}$ cut off from $T_{1}$ two equal prisms $Q_{1} Q_{4} P_{4} P_{1} D C, Q_{2} Q_{3} P_{3} P_{2} K L$ with a volume of $\left(a-\frac{a}{\sqrt{2}}\right)^{2} \cdot \frac{a}{2}$ and 4 equal pyramids $F P_{1} P_{2} N, F P_{3} P_{4} M, E Q_{1} Q_{2} A, E Q_{3} Q_{4} B$ with a volume of $\frac{1}{3}\left(\frac{a}{\sqrt{2}}\right)^{2} \cdot \frac{a}{4}$. Therefore, the volume of the common part of the two cubes $T_{1}$ and $T_{2}$ is:

$$
a^{3}-a\left(a-\frac{a}{\sqrt{2}}\right)^{2}-\frac{a}{3}\left(\frac{a}{\sqrt{2}}\right)^{2}=\frac{a^{3}(3 \sqrt{2}-2)}{3}
$$

Answer: $\frac{a^{3}(3 \sqrt{2}-2)}{3}$
11.196 Two equal cubes with edge a have a common segment $E F$, the ends of which are the midpoints of two opposite edges, not belonging to the same face. One of the cubes is obtained by rotating the other around the line $\boldsymbol{E F}$ by $90^{\circ}$. Find the volume of the common part of these cubes.