It is best to remember the following conclusions about regular tetrahedron:

Regular tetrahedron with side length A: (1) high position a√6/3 (that is, three-thirds of the roots are six times that of A)-this conclusion is more convenient to use in this problem;

(2) The catching radius is a√6/4 (that is, six times a of the quarter root)

(3) The radius of the inner contact surface is a√6/ 12 (that is, the root sign of one twelfth is six times that of a).

From the conclusion (1), we can see that the height of the regular tetrahedron in this question is √6/3, so its volume is.

( 1/3)( √3/4)( √6/3)= √2/ 12

Then the ABC at the bottom is actually a spherical cap, and the volume formula of the spherical cap is as follows (this knowledge point should not be within the scope):

V= h (2/3) R 2 where h is the height of the spherical cap and r is the radius of the ball where the spherical cap is located.

The height of the spherical cap = the radius of the ball-the height of the regular tetrahedron =1-6/3.

So the volume of the spherical cap = (1-√ 6/3) (2/3) = (6-2 √ 6)/9.

So the total volume =√2/ 12+ √ (6-2√6)/9.