1. Let $a$ and $b$ be skew lines, and on line $a$ there are three points $A$, $B$, and $C$, such that $AB = BC$. Through $A$, $B$, and $C$, perpendiculars to line $b$ are drawn, intersecting $b$ at points $D$, $E$, and $F$ respectively. It is given that $AD = \sqrt{15}$, $BE = \frac{7}{2}$, and $CF = \sqrt{10}$. Then the distance between the skew lines $a$ and $b$ is $\qquad$.
$=1 . \sqrt{6}$.
Let the distance between the skew lines $a$ and $b$ be $x$. Construct the common perpendicular segment $MN$ of lines $a$ and $b$, with $N \in a$. Through point $M$, draw line $a' \parallel a$, and let lines $b$ and $a'$ determine plane $a$. By the given conditions, we know $MN = x$, and $AB = BC$, then
$$
2 \sqrt{B E^{2}-x^{2}}=\sqrt{A D^{2}-x^{2}}+\sqrt{C F^{2}-x^{2}} \text {. }
$$

Solving for $x$ yields $x=\sqrt{6}$.