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The base of a right triangular prism ABCA1B1C1 is an isosceles right triangle with legs \( AC = BC = a \). The vertices \( M \) and \( N \) of a regular tetrahedron \( MN P Q \) lie on the line CA1, and the vertices \( P \) and \( Q \) lie on the line AB1. Find: a) the volume of the prism; b) the distance between the midpoints of segments \( MN \) and \( PQ \).
Since the opposite edges $M N$ and $P Q$ of the regular tetrahedron are perpendicular, the lines $C A_1$ and $A B_1$ containing them (Fig.1) are also perpendicular. Since $B_1 C_1 \perp A_1 C_1$ and $B_1 C_1 \perp C C_1$, $B_1 C_1$ is perpendicular to the plane $A A_1 C_1 C$. Since $A C_1$ is the orthogonal projection of the inclined line $A B_1$ on the plane $A A_1 C_1 C$, by the theorem of three perpendiculars, $A C_1 \perp C A_1$. Therefore, the rectangle $A A_1 C_1 C$ is a square. Thus, the lateral edges of the given prism are equal to $a$. Therefore,

$$
V_{A B C A} 1 B_1 C_1 = S_{\triangle A B C} \cdot A A_1 = \frac{1}{2} a^2 \cdot a = \frac{1}{2} a^3.
$$

The line $A_1 C$ is perpendicular to the plane $A B_1 C_1$, since it is perpendicular to the intersecting lines $B_1 C_1$ and $A C_1$ of this plane. Let $F$ be the center of the square $A A_1 C_1 C$. Then the perpendicular $F H$, dropped from point $F$ to the line $A B_1$, is the common perpendicular of the skew lines $C A_1$ and $A B_1$. Consider the right triangle $A B_1 C_1$ (Fig.2). Denote $\angle B_1 A C_1 = \alpha$. Then

$$
\operatorname{tg} \alpha = \frac{B_{C_{C}} C_1}{A C_1} = \frac{1}{\sqrt{2}}, \cos \alpha = \sqrt{\frac{2}{3}}, \sin \alpha = \frac{1}{\sqrt{3}}
$$

$$
F H = A F \sin \angle F A H = \frac{a}{\sqrt{2}} \cdot \sin \alpha = \frac{a}{\sqrt{2}} \cdot \frac{1}{\sqrt{3}} = \frac{a}{\sqrt{6}}.
$$

Since the segment connecting the midpoints of opposite edges of a regular tetrahedron is the common perpendicular of the lines containing these edges, the length $F H$ is the required distance.

## Answer

J) $\left.\frac{a^{\frac{3}{x}}}{2} ; A\right) \frac{a}{\sqrt{6}}$.