Find the equation of the circle circumscribing the triangle formed by the three points $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$. Assuming the equation of the circle circumscribing the triangle formed by the three given points is given by the sphere through the three points and the plane through the three points.  
Plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
and assuming the equation of the sphere is $x^2+y^2+z^2+2ux+2vy+2wz+d=0$.  
I've not been able to find the value for $d$, putting the three points in the equation of sphere I got $3$ equations with $4$ variables.

Given the points
$$
\cases{
A=(a,0,0)\\
B=(0,b,0)\\
C=(0,0,c)\\
p = (x,y,z)
}
$$
the sphere center $p_0$ is determined by the intersection of plane $\Pi$ with the line $L$ such that
$$
\cases{
\Pi\to (p-A)\cdot\vec n=0\\
\vec n = (A-B)\times(B-C)\\
L\to \left((p-\frac{A+B}{2})\cdot(A-B)=0\right)\cap \left((p-\frac{A+C}{2})\cdot(A-C)=0\right)\\
L\to p=(0,\frac{b^2-a^2}{2b},\frac{c^2-a^2}{2c})+\lambda(1,\frac ab,\frac ac)
}
$$
then
$$
p_0 = \frac{1}{2(a^2b^2+a^2c^2+b^2 c^2)}\left(a^3 \left(b^2+c^2\right),b^3 \left(a^2+c^2\right),c^3\left(a^2+b^2\right)\right)
$$
and
$$
r = \|p_0-A\| = \frac{1}{2} \sqrt{\frac{\left(a^2+b^2\right) \left(a^2+c^2\right) \left(b^2+c^2\right)}{a^2b^2+a^2c^2+b^2 c^2}}
$$
Attached a plot for $a=1,b=2,c=3$. In pink the plane $(p-A)\cdot \vec n=0$. In green the plane $(p-\frac{A+B}{2})\cdot(A-B)=0$ and in yellow the plane $(p-\frac{A+C}{2})\cdot(A-C)=0$. In red the line $p=(0,\frac{b^2-a^2}{2b},\frac{c^2-a^2}{2c})+\lambda(1,\frac ab,\frac ac)$

