1. In $\triangle A B C$, it is known that the height on side $B C$ is 3, and two lines $l_{1} 、 l_{2}$ parallel to $B C$ divide the area of $\triangle A B C$ into three equal parts. Then the distance between the lines $l_{1}$ and $l_{2}$ is
$-1 . \sqrt{6}-\sqrt{3}$.

As shown in Figure 2, let $l_{1}$ and $l_{2}$ intersect the altitude from $A$ to side $BC$ at points $E$ and $F$, respectively, intersect $AB$ at points $G$ and $J$, and intersect $AC$ at points $H$ and $K$.
By similar triangles, we have
$$
\frac{A E}{A D}=\sqrt{\frac{S_{\triangle 1 G H}}{S_{\triangle A B C}}}=\frac{1}{\sqrt{3}} \text {. }
$$

Since $A D=3$, we have $A E=\sqrt{3}$.
Similarly, $A F=\sqrt{6}$.
Therefore, $E F=\sqrt{6}-\sqrt{3}$.