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
p2 = b2+c2
q2 = c2+a2
r2 = a2+b2
K = sqrt((p+q+r)(p+q-r)(p-q+r)(-p+q+r))/4

multiplying out the four factors gives

K = sqrt(2q2r2+2r2p2+2p2q2-p4-q4-r4)/4

substituting for p2, q2, r2 gives

K = sqrt(2(c2+a2)(a2+b2)+2(a2+b2)(b2+c2)+2(b2+c2)(c2+a2)-(b2+c2)2-(c2+a2)2-(a2+b2)2)/4

multiplying out all these pairs of factors and collecting terms gives

K = sqrt(4b2c2+4c2a2+4a2*b2)/4
K = sqrt(b2c2+c2a2+a2b2)/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


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,


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(A2+B2+C2), or


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

(1/3)KH = (1/3) * sqrt(b2c2+c2a2+a2*b2)/2 * abc/sqrt(b2c2+c2a2+a2*b2)

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!