Find the volume of the tetrahedron $ABCD$ with edges $AB=3$, $AC=5$, and $BD=7$, if the distance between the midpoints $M$ and $N$ of its edges $AB$ and $CD$ is 2, and the line $AB$ forms equal angles with the lines $AC$, $BD$, and $MN$.
Complete the tetrahedron $ABCD$ to a parallelepiped $APBQECGD$ by drawing pairs of parallel planes through its opposite edges ($AE\|PC\|BG\|QD$). Then $AE \| MN, AE=MN=2$. On the extension of segment $CE$ beyond point $E$, lay off segment $EF$ equal to $CE$. Then $AFEP$ is a parallelogram, so $AF\|PE\|BD$ and $AF=PE=BD$

$=7$. Since the line $AB$ forms equal angles with the lines $BD$ and $MN$, it forms equal angles with the

parallel lines $AF$ and $AE$. Therefore, the line $AB$ forms equal angles with the lines $AF, AE$, and $AC$,

lying in the plane $AEC P$. Therefore, the orthogonal projection of point $B$ onto this plane falls on

the bisector of each of the angles between the lines $AF$ and $AE$, $AF$ and $AC$, $AE$ and $AC$. Consequently, the orthogonal projection of point $B$ onto the plane $AEC P$ coincides with point $A$, i.e., $AB$ is perpendicular to this plane. Thus, $AB$ is the height of the quadrilateral pyramid $BAECP$, the base of which is the parallelogram $AEC P$ with sides $AE=CP=2$ and diagonals $AC=5$ and $EP=7$. On the extension of edge $AE$ beyond point $E$, lay off segment $EH=AE=2$.

Then $AH=2AE=4, CH=PE=7$, and

$$
S_{AEC P}=S_{\triangle AHC}=\sqrt{8(8-7)(8-5)(8-4)}=4 \sqrt{6}
$$

Note that both the volume of the tetrahedron $ABCD$ and the volume of the pyramid $BAECP$ are one-third of the volume of the parallelepiped $APBQECGD$. Therefore,

$$
V_{ABCD}=V_{BAECP}=\frac{1}{3} S_{AEC P} \cdot AB=\frac{1}{3} \cdot 4 \sqrt{6} \cdot 3=4 \sqrt{6}
$$

## Answer

$4 \sqrt{6}$.

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