6. As shown in Figure 1, in the unit cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$, $E$, $F$, and $G$ are the midpoints of edges $A A_{1}$, $C_{1} D_{1}$, and $D_{1} A_{1}$, respectively. Then the distance from point $B_{1}$ to the plane containing $\triangle E F G$ is . $\qquad$
6. $\frac{\sqrt{3}}{2}$.

Solution 1 As shown in Figure 3, the plane passing through points $E$, $F$, and $G$ intersects the cube to form a regular hexagon. It is easy to see that this plane perpendicularly bisects the diagonal $B_{1} D$ of the cube.

Since $B_{1} D=\sqrt{3}$, the distance from point $B_{1}$ to the plane $E F G$ is $\frac{\sqrt{3}}{2}$.
Solution 2 It is easy to know that $S_{\triangle B_{1} F G}=\frac{3}{8}$.
The distance from point $E$ to the plane $B_{1} F G$ is $h_{0}=\frac{1}{2}$, so
$V_{E-B_{1} F C}=\frac{1}{3} h_{0} S_{\triangle B_{1} F G}=\frac{1}{16}$.
Also, $E F=\frac{\sqrt{6}}{2}$, $G F=G E=\frac{\sqrt{2}}{2}$, then
$S_{\triangle E G F}=\frac{\sqrt{3}}{8}$.
Therefore, the distance from point $B_{1}$ to the plane $E F G$ is
$$
\frac{3 V_{B_{1}-E G F}}{S_{\triangle E G F}}=\frac{\sqrt{3}}{2} .
$$