3. For the tetrahedron $S-ABC$ with three lateral edges mutually perpendicular, $O$ is a point inside the base $\triangle ABC$. Then, the minimum value of $W=\tan \angle OSA \cdot \tan \angle OSB \cdot \tan \angle OSC$ is
3. $2 \sqrt{2}$

Analysis: According to the problem, $\cos ^{2} \angle O S A+\cos ^{2} \angle O S B+\cos ^{2} \angle O S C=1$,

thus $\sin ^{2} \angle O S C=\cos ^{2} \angle O S A+\cos ^{2} \angle O S B \geqslant 2 \cos \angle O S A \cdot \cos \angle O S B$, similarly, there are two more inequalities. Multiplying the three inequalities, we get $W \geq 2 \sqrt{2}$.