3. In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, points $E, F, G$ are the midpoints of edges $A A_{1}, C_{1} D_{1}, D_{1} A_{1}$, respectively. Then the distance from point $B_{1}$ to plane $E F G$ is $\qquad$ .
3. $\frac{\sqrt{3}}{2}$.
(Method 1) As shown in Figure 2, the plane passing through $E, F, G$ intersects the cube, forming a regular hexagon. It is easy to see that this plane perpendicularly bisects the cube's diagonal $B_{1} D$, and since $B_{1} D=\sqrt{3}$, the distance from $B_{1}$ to the plane $E F G$ is $h=\frac{\sqrt{3}}{2}$.
(Method 2) It is easy to know that
$$
S_{\triangle D_{1} F G}=1-S_{\triangle B_{1} A_{1} G}-S_{\triangle B_{1} C_{1} F}-S_{\triangle D_{1} F G}=1-\frac{1}{4}-\frac{1}{4}-\frac{1}{8}=\frac{3}{8},
$$

and the distance from point $E$ to the plane $B_{1} F G$ is $h_{0}=\frac{1}{2}$, so
$$
V_{\text {tetrahedron } E B_{1} F G}=\frac{1}{3} h S_{\triangle B_{1} F G}=\frac{1}{16} .
$$

Also,
$$
E F^{2}=E A_{1}^{2}+A_{1} F^{2}=E A_{1}^{2}+\left(A_{1} D_{1}^{2}+D_{1} F^{2}\right)=\frac{1}{4}+1+\frac{1}{4}=\frac{3}{2},
$$

which means $E F=\frac{\sqrt{6}}{2}, G F=G E=\frac{\sqrt{2}}{2}$,
$$
\cos \angle E G F=\frac{G E^{2}+G F^{2}-E F^{2}}{2 G E \cdot G F}=-\frac{1}{2},
$$

so $\angle E G F=120^{\circ}$, then
$$
S_{\triangle E G F}=\frac{1}{2} G E \cdot G F \sin 120^{\circ}=\frac{\sqrt{3}}{8},
$$

If the distance from $B_{1}$ to the plane $E F G$ is $h$, then
$$
\frac{1}{16}=V_{\text {tetrahedron } E B_{1} F G}=\frac{1}{3} h \cdot S_{\triangle E G F}=\frac{\sqrt{3}}{24} h,
$$

so $h=\frac{\sqrt{3}}{2}$.