12.3.7 * 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 the points $S, A, B, C$ all lie on a sphere with center $O$. Then the distance from point $O$ to the plane $ABC$ is $\qquad$.
As shown in the figure, in the isosceles right triangle $A B C$, the midpoint of the hypotenuse $A B$ is $D$, then $D$ is the circumcenter of $\triangle A B C$. Given $S A=S B=S C=2$, it is known that the projection of $S$ on the base $A B C$ is $D$, thus, the center of the sphere $O$ lies on $S D$, and $O A=O B=O S$. Therefore, $O$ is the circumcenter of $\triangle S A B$. Since $\triangle S A B$ is an equilateral triangle, $S D=\sqrt{2^{2}-1^{2}}=\sqrt{3}$, so the distance from $O$ to the plane $A B C$ is $O D-\frac{1}{3} \cdot S D=\frac{\sqrt{3}}{3}$.