12. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with hypotenuse $AB$, and $SA=SB=SC=AB=2$. Suppose points $S, A, B, C$ all lie on a sphere with center $O$. Then the distance from point $O$ to the plane $ABC$ is $\qquad$
12. $\frac{\sqrt{3}}{3}$.

As shown in Figure 3.
Since $S A=S B=S C$, the projection of point $S$ on the plane $A B C$ is the circumcenter of $\triangle A B C$, which is the midpoint $H$ of $A B$.

Similarly, the projection of point $O$ on the plane $A B C$ is also $H$. Therefore, we only need to find the length of $O H$ in the equilateral $\triangle S A B$, i.e., $O H=\frac{1}{3} S H=\frac{1}{3} \times 2 \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{3}$.