[ Distance between skew lines]

On the line $l$ in space, points $A, B$ and $C$ are sequentially located, with $A B=18$ and $B C=14$. Find the distance between the lines $l$ and $m$, if the distances from points $A, B$ and $C$ to the line $m$ are 12, 15 and 20, respectively.

#
Let $A 1, B 1$ and $C 1$ be the feet of the perpendiculars dropped from points $A, B$ and $C$ to the line $m$ respectively (Fig.1). According to the problem,

$$
A A 1=12, B B 1=15, C C 1=20 .
$$

Assume that lines $l$ and $m$ lie in the same plane. It is clear that they cannot be parallel. If the intersection point of lines $l$ and $m$ lies outside the segment $A 1 C 1$ (Fig.2), then by dropping perpendiculars $A D$ and $B E$ from points $A$ and $B$ to lines $B B 1$ and $C C 1$ respectively, we obtain similar triangles $A D B$ and $B E C$, which is impossible, since

$$
\frac{B D}{A B}=\frac{3}{18}=\frac{1}{6} \neq \frac{5}{14}=\frac{C E}{B C}
$$

Similarly for the case when the intersection point of lines $l$ and $m$ lies on the segment $A 1 C 1$. Thus, $l$ and $m$ are skew lines. Consider the orthogonal projection of lines $l$ and $m$ onto the plane $\alpha$, perpendicular to the line $m$ (Fig.3). Let points $P, Q$ and $R$ be the orthogonal projections of points $A$, $B$ and $C$ respectively, and $M$ be the projection of points $A 1, B 1$ and $C 1$. By the property of parallel projection, and since segments $A A 1, B B 1$ and $C C 1$ are perpendicular to the line $m$, they are parallel to the plane $\alpha$. Therefore,

$$
P M=A A 1=12, Q M=B B 1=15, M R=C C 1=20 .
$$

Let $P Q=9 x, Q R=7 x, \angle M R P=\phi$. From triangles $M Q R$ and $M P R$ using the cosine theorem, we find that

$$
\cos \phi=\frac{400+47 x^{2}-225}{2 \cdot 20 \cdot 7 x}
$$

$$
\cos \phi=\frac{400+256 x^{2}-144}{2 \cdot 20 \cdot 16 x} .
$$

From the equation

$$
\frac{400+47 x^{2}-225}{2 \cdot 20 \cdot 7 x}=\frac{400+256 x^{2}-144}{2 \cdot 20 \cdot 16 x}
$$

we find that $x=1$. Therefore, $P R=16 x=16$. Since $M P 2+P R 2=144+256=400=M R 2$, triangle $M P R$ is a right triangle, with $\angle M P R=90^{\circ}$. Therefore, $P M$ is perpendicular to the projection $P R$ of the inclined line $l$ on the plane $\alpha$. By the theorem of three perpendiculars, $M P \perp l$, and since the line $P M$ lies in the plane $\alpha$, perpendicular to the line $m$, then $M P \perp m$. It remains to note that the segment $MP$ is the orthogonal projection on the plane $\alpha$ of the common perpendicular of lines $l$ and $m$. Since this common perpendicular is parallel to the plane $\alpha$, it is equal to the segment $MP$. Therefore, the distance between the lines $l$ and $m$ is 12.

## Answer

12.00