2. Given four non-coplanar points $A, B, C, D$, where the distance between any two points is distinct, consider a plane $\alpha$ with the following property: three of the points $A, B, C, D$ are at the same distance from $\alpha$, and the fourth point is at a distance that is twice the distance of one of the three points from $\alpha$. The number of such planes $\alpha$ is $\qquad$
2. Solution: Taking any one of the four points as the fourth point (the distance to the plane $\alpha$ is twice the distance of the other points to $\alpha$), there are 4 ways to choose. When $D$ is the fourth point, the distances from $A, B, C$ to $\alpha$ are equal, (1) if the three points are on the same side of $\alpha$, there are two cases depending on whether $D$ is on the same side or the opposite side of the three points; (2) if the three points are on different sides of $\alpha$, there are 3 cases, and for each case, $D$ can be on the same side as two of the points or on the same side as one of the points, each with two scenarios, so there are 6 cases in this situation. Therefore, when $D$ is the fourth point, there are 8 cases in total, hence there are 32 cases in total.