III OM - I - Task 4

a) Given points $ A $, $ B $, $ C $ not lying on a straight line. Determine three mutually parallel lines passing through points $ A $, $ B $, $ C $, respectively, so that the distances between adjacent parallel lines are equal.
b) Given points $ A $, $ B $, $ C $, $ D $ not lying on a plane. Determine four mutually parallel planes passing through points $ A $, $ B $, $ C $, $ D $, respectively, so that the distances between adjacent parallel planes are equal.
a) Suppose that the lines $a$, $b$, $c$ passing through points $A$, $B$, $C$ respectively and being mutually parallel satisfy the condition of the problem, that is, the distances between adjacent parallel lines are equal. Then the line among $a$, $b$, $c$ that lies between the other two is equidistant from them. Let this line be, for example, line $b$. In this case, points $A$ and $C$ are equidistant from line $b$ and lie on opposite sides of it; therefore, line $b$ intersects segment $AC$ at its midpoint $M$. From the fact that points $A$, $B$, $C$ do not lie on a straight line, it follows that point $M$ is different from point $B$.
Hence the construction: we draw line $b$ through point $B$ and through the midpoint $M$ of segment $AC$, and then we draw through points $A$ and $C$ lines $a$ and $c$ parallel to line $b$ (Fig. 12). The parallel lines $a$, $b$, $c$ determined in this way solve the problem, since points $A$ and $C$, and thus lines $a$ and $c$, are equidistant from line $b$ and lie on opposite sides of this line.
We found the above solution assuming that among the sought lines $a$, $b$, $c$, the line $b$ lies between lines $a$ and $c$; since the "inner" line can equally well be $a$ or $c$, the problem has three solutions.
b) Suppose that the planes $\alpha$, $\beta$, $\gamma$, $\delta$, passing through points $A$, $B$, $C$, $D$ respectively and being mutually parallel, satisfy the condition of the problem, that is, the distances between adjacent planes are equal. Let these planes be in the order $\alpha$, $\beta$, $\gamma$, $\delta$. We mean by this that plane $\beta$ is equidistant from planes $\alpha$ and $\gamma$, and plane $\gamma$ is equidistant from planes $\beta$ and $\delta$.
In this case, points $A$ and $C$ are equidistant from plane $\beta$ and lie on opposite sides of it, so plane $\beta$ passes through the midpoint $M$ of segment $AC$. Similarly, plane $\gamma$ passes through the midpoint $N$ of segment $BD$. From the fact that points $A$, $B$, $C$, $D$ do not lie in a plane, it follows that point $M$ is different from point $B$, and point $N$ is different from point $C$.
From this, we derive the following construction. We connect point $B$ with the midpoint $M$ of segment $AC$, and point $C$ with the midpoint $N$ of segment $BD$ (Figure 13 shows a parallel projection of the figure). Lines $BM$ and $CN$ are skew; if they lay in the same plane, then points $A$, $B$, $C$, $D$ would lie in the same plane, contrary to the assumption. We know from stereometry that through two skew lines $BM$ and $CN$ one can draw two and only two parallel planes $\beta$ and $\gamma$.
To do this, we draw through point $M$ a line $m$ parallel to line $CN$, and through point $N$ a line $n$ parallel to line $BM$; plane $\beta$ is then determined by lines $m$ and $BM$, and plane $\gamma$ by lines $n$ and $CN$.
Finally, we draw through points $A$ and $D$ planes $\alpha$ and $\delta$ parallel to planes $\beta$ and $\gamma$; we can determine them, as indicated in Figure 13, by drawing through each of points $A$ and $D$ lines parallel to lines $BM$ and $CN$.
The planes $\alpha$, $\beta$, $\gamma$, $\delta$ determined in this way solve the problem, since points $A$ and $C$, and thus planes $\alpha$ and $\gamma$, are equidistant from plane $\beta$ and lie on opposite sides of this plane - and similarly, planes $\beta$ and $\delta$ are equidistant from plane $\gamma$ and lie on opposite sides of this plane.
We found the above solution assuming that the sought planes are in the order $\alpha$, $\beta$, $\gamma$, $\delta$. For other orders of the sought planes, we will find other solutions in the same way. The number of all possible orders, or permutations of the letters $\alpha$, $\beta$, $\gamma$, $\delta$, is $4!$, i.e., $24$. However, note that two "reverse" permutations, such as $\alpha$, $\beta$, $\gamma$, $\delta$ and $\delta$, $\gamma$, $\beta$, $\alpha$, give the same solution. Therefore, the problem has $\frac{24}{2} = 12$ solutions corresponding to the permutations.