2. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with $AB$ as the hypotenuse, $SA=SB=SC=2$, $AB=2$, and suppose points $S, A, B, C$ all lie on a sphere with center $Q$, then the distance from point $O$ to the plane $ABC$ is $\qquad$ (1997 National High School Competition Question)
2. $\frac{\sqrt{3}}{3}$. In the isosceles right triangle $A B C$, let the midpoint of the hypotenuse $A B$ be $D$, then $D$ is the circumcenter of $\triangle A B C$. Given $S A=S B=S C=2$, we know that the projection of $S$ on the base $A B C$ is $D$, and the center of the sphere is $O$. Since $\triangle S A B$ is an equilateral triangle, then $S D=\sqrt{3}$. Therefore, the distance from $O$ to the plane $A B C$ is $O D=\frac{\sqrt{3}}{3}$.