POSSIBLE SOLUTION

The key to this solution is to place the final tetrahedron inside a
rectangular box (x by y by z) so that its vertices O, P, Q, and R are also vertices of the box and so that half-side lengths of the original triangle are the side lengths of the box. See the figure below. In particular, P, Q, and R are the midpoints of AB, BC, and AC respectively, and O is the coincidence of A, B, and C.

The box is partitioned by the desired tetrahedron into five tetrahedra: the one we're interested in, and four other congruent ones each of which has base area xy/2 and height z, and hence volume (1/3)(xy/2) = (1/6)(x y z). This gives us the volume of OPQR as x y z - 4 (1/6)(x y z) = (1/3)(x y z). So, all we have to do is determine x, y, and z and we're done.

The side lengths x, y, and z are determined by the system
x^2 + y^2 = (21/2)^2
x^2 + z^2 = (10)^2
y^2 + z^2 = (11/2)^2
which yields x^2 = 90, y^2 = 81/4, and z^2 = 10 so that x = 3 sqrt(10), y = 9/2, and z = sqrt(10) so that the volume of the tetrahedron is (1/3) x y z = 45.