Find the volume of the tetrahedron with verticies (1,1,2), (3,-4,6), (-7,0,-1) and (-1,5,8).

Assume A(1,1,2) B(3,-4,6) and C(-7,0,-1) will form the base of the tetrahedron

So vector(AB) = (2,-5,4) and
vector(AC) = (-8,-1,-3)

the area of the parallelogram is the magnitude of the cross product of those two vectors, so the trianle would be 1/2 of that
AB x AC = (19,-26,-42)
and |(19,-26,-42)| = √2801
so the area of the triangular base is
(1/2)√2801

Since (19,-26,-42) is also a normal to the plane of the base, we can find the equation of that readily
19x - 26y - 42z + c = 0
using the point (-7,0,-1)
19(-7) - 1(-42) + c = 0
c = 91
equation of plance
19x - 26y - 42z + 91 = 0

so by taking the distance from our fourth point (-1,5,8) to that base we have the height of the tetrahedron

height = |-1(19) + 5(-26) + 8(-42) + 91|/√(19^2+(-26)^2+(-42)^2)
= 394/√2801

volume of a tetrahedron
= (1/3)base x height
= (1/3)(1/2(√2801 (394/√2801
= 197/3 cubic unit