Three, pairwise perpendicular lines intersect at a point $O$. The points on one line that are one unit away from $O$ are $X_{1}$ and $X_{1}^{\prime}$, and the points that are two units away are $X_{2}$ and $X_{2}^{\prime}$. Similarly, we denote the points on the other two lines as $Y_{1}, Y_{1}^{\prime}, Y_{2}, Y_{2}^{\prime}, Z_{1}, Z_{1}^{\prime}, Z_{2}, Z_{2}^{\prime}$. - We form planes through every possible combination of a 1-indexed point and a 2-indexed point on each of the other lines, for example, through $X_{1}^{\prime}$, $Y_{2}$, and $Z_{2}^{\prime}$.

Determine the volume of the solid enclosed by these planes.
We choose the notation such that each point without a comma lies on one of the rays originating from $O$, and the points with a comma lie on the other.

The 8 planes determined by the 2nd index points of the three lines form a regular octahedron $T_{2}$, whose volume is $32 / 3$. By taking $Z_{1}$ instead of $Z_{2}$ and $Z_{1}^{\prime}$ instead of $Z_{2}^{\prime}$ in these planes, the 8 planes form a body $T_{2}$ that can also be derived from $T_{2}$ by a contraction along the $z$-axis, so its volume is $16 / 3$. Similarly, the bodies $T_{x}, T_{y}$, and $T_{z}$, formed by contraction along the $x$ and $y$ axes, respectively, together with the faces of $T_{z}$, bound the body $T$ under investigation, which has 24 faces (Figure 1).
![](https://cdn.mathpix.com/cropped/2024_05_02_f314f4b07c42e50c53dfg-1.jpg?height=600&width=1276&top_left_y=464&top_left_x=411)

Figure 1

Each of $T_{x}, T_{y}$, and $T_{z}$ is symmetric with respect to the planes determined by the three axes taken in pairs, so the same is true for their common part, $T$. Therefore, it is sufficient to examine the part of $T$ that falls into one of the 8 regions defined by the 3 planes. Let this be the part that falls into the tetrahedron $O X_{2} Y_{2} Z_{2}$.

The plane $X_{2} Y_{2} Z_{1}$ divides the tetrahedron $O X_{2} Y_{2} Z_{2}$ into two parts of equal volume (since the non-common vertices $O, Z_{2}$ of the parts are at equal distances from the intersecting plane), and it intersects the faces $O Z_{2} Y_{2}$ and $O Z_{2} X_{2}$ along their medians $Y_{2} Z_{1}$ and $X_{2} Z_{1}$, respectively. The planes $X_{2} Y_{2} Z_{1}$ and $Y_{2} Z_{2} X_{1}$ intersect at point $Y_{2}$, and they intersect the face $O X_{2} Z_{2}$ along its medians $X_{2} Z_{1}$ and $X_{1} Z_{2}$, respectively, so the centroid $S_{2}$ of this face is also on both planes. The intersection line of the planes $X_{2} Y_{2} Z_{1}$ and $Y_{2} Z_{2} X_{1}$ is therefore the median $Y_{2} S_{2}$ of the tetrahedron $O X_{2} Y_{2} Z_{2}$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_f314f4b07c42e50c53dfg-1.jpg?height=566&width=1192&top_left_y=1544&top_left_x=453)

Figure 2

Similarly, we find that the other intersection lines are the medians $Z_{2} S_{3}$ and $X_{2} S_{1}$, where $S_{3}$ is the centroid of the face $O X_{2} Y_{2}$, and $S_{1}$ is the centroid of the face $O Y_{2} Z_{2}$. Thus, the intersection point of the planes $X_{2} Y_{2} Z_{1}$, $Y_{2} Z_{2} X_{1}$, and $Z_{2} X_{2} Y_{1}$ is the centroid $S$ of the tetrahedron $O X_{2} Y_{2} Z_{2}$. The part of $T$ that falls into this tetrahedron consists of the union of pyramids with apex $O$ and bases

$$
S_{1} S S_{2} Z_{1}, \quad S_{2} S S_{3} X_{1}, \quad S_{3} S S_{1} Y_{1}
$$

Since the area of the triangle $X_{2} Z_{2} Y_{1}$ is 1/6 of the area of the quadrilateral $S_{3} S S_{1} Y_{1}$ (because the area of the triangle $Y_{1} S S_{3}$ is half the area of the triangle $X_{2} S S_{3}$, which is 1/4 of the area of the triangle $X_{2} Z_{2} S_{3}$, which is 2/3 of the area of the triangle $X_{2} Z_{2} Y_{1}$), the volume of the pyramid $O S_{3} S S_{1} Y_{1}$ is 1/6 of the volume of the tetrahedron $O X_{2} Z_{2} Y_{1}$, or 1/12 of the volume of the tetrahedron $O X_{2} Y_{2} Z_{2}$. Therefore, the combined volume of the above three pyramids is 1/4 of the volume of the tetrahedron $O X_{2} Y_{2} Z_{2}$, and thus the volume of $T$ is also one-fourth of the volume of $T_{2}$:

$$
V_{T}=\frac{1}{4} V_{2}=\frac{1}{4} \cdot \frac{32}{3}=\frac{8}{3} \quad \text { cubic units. }
$$

Remark. The body under investigation is a deltoidic tetrahedron (deltoid icositetrahedron) of the regular (cubic) crystal system. Its general notation is $h, k, k$, where $h$ and $k (h < k)$ are the different segments cut from the axes by its planes, in our case $1,2,2$.