Example 10 (1998 National High School League Question) In a cube, among the 8 vertices, the midpoints of the 12 edges, the centers of the 6 faces, and the center of the cube, a total of 27 points, the number of groups of three collinear points is ( ).
A. 57
B. 49
C. 43
D. 37
Solution: Choose B. Reason: Perform scenario segmentation and reduction. The number of collinear triplets with both ends as vertices is $\frac{8 \cdot 7}{2}=28$; The number of collinear triplets with both ends as the centers of faces is $\frac{6 \cdot 1}{2}=3$; The number of collinear triplets with both ends as the midpoints of edges is $\frac{12 \cdot 3}{2}=18$, and there are no other types of collinear triplets, so the total number of collinear triplets is $28+3+18=49$.