The question goes like this:

Let (a,0,0), (0,b,0), (0,0,c) be three vertices of a triangle. Now, there are two parts to this question: First, we ask, what is the area of this triangle? Next we ask, what is the volume of the tetrahedron whose vertices are those of this triangle and the origin?

--- PART 1 ---

What is the area of that triangle?

If we let p,q,r be the sides of the triangle, then Heron's formula gives

- K = sqrt(s(s-p)(s-q)(s-r)) = sqrt((p+q+r)(p+q-r)(p-q+r)(-p+q+r))/4

- p
^{2}= b^{2}+c^{2} - q
^{2}= c^{2}+a^{2} - r
^{2}= a^{2}+b^{2}

- K = sqrt((p+q+r)(p+q-r)(p-q+r)(-p+q+r))/4

multiplying out the four factors gives

- K = sqrt(2q
^{2}r^{2}+2r^{2}p^{2}+2p^{2}q^{2}-p^{4}-q^{4}-r^{4})/4

substituting for p^{2}, q^{2}, r^{2} gives

- K = sqrt(2(c
^{2}+a^{2})(a^{2}+b^{2})+2(a^{2}+b^{2})(b^{2}+c^{2})+2(b^{2}+c^{2})(c^{2}+a^{2})-(b^{2}+c^{2})^{2}-(c^{2}+a^{2})^{2}-(a^{2}+b^{2})^{2})/4

multiplying out all these pairs of factors and collecting terms gives

- K = sqrt(4b
^{2}c^{2}+4c^{2}a^{2}+4a^{2}*b^{2})/4 - K = sqrt(b
^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2})/2

--- PART 2 ---

Now, we know the volume of the tetrahedron is (1/6)abc. Let's find the volume using triangle (a,0,0), (0,b,0), (0,0,c) as a base. The equation of a plane is

- Ax+By+Cz=D,

and the intercepts are a=D/A, b=D/B, and c=D/C. So if we let D=abc, then we get A=bc, B=ac, and C=ab, giving us the equation of the plane,

- bcx+cay+abz=abc,

and the normal vector

- <bc, ca, ab>

The distance, H, of a plane Ax+By+Cz=D from the origin is simply H=D/sqrt(A^{2}+B^{2}+C^{2}), or

- H=abc/sqrt(b
^{2}c^{2}+c^{2}a^{2}+a^{2}*b^{2})

The volume of the tetrahedron with base area K and height H is

- (1/3)KH = (1/3) * sqrt(b
^{2}c^{2}+c^{2}a^{2}+a^{2}*b^{2})/2 * abc/sqrt(b^{2}c^{2}+c^{2}a^{2}+a^{2}*b^{2})

which indeed equals (1/6)abc. If we hadn't calculated K earlier using Heron's formula and a heck of a lot of collecting of terms, we could have figured it out (more easily, I dare say) using the formula for the plane, its distance from the origin, and the volume of the tetrahedron. It's nice to know the two methods agree!