12. Place 1 sphere with a radius of 2 between two inclined planes whose lines of intersection are on a horizontal plane, with the angles between these planes and the horizontal plane being $60^{\circ}$ and $30^{\circ}$, respectively. Then the distance from the center of the sphere to the horizontal plane is $\qquad$
12. $\sqrt{3}+1$

Passing through the center of the sphere $O$, a plane perpendicular to the horizontal plane is made, resulting in the cross-sectional diagram as shown on the right. It is easy to get $\angle O C F=\frac{180^{\circ}-60^{\circ}-30^{\circ}}{2}=$ $45^{\circ}, \angle E C O=30^{\circ}+45^{\circ}=75^{\circ}$

Therefore, $O C=2 \sec 45^{\circ}=2 \sqrt{2}$, so $O E=O C$. $\sin 75^{\circ}=2 \sqrt{2}\left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\right)=\sqrt{3}+1$.