Fixed Sphere tangent to a plane

Question

Let $A(0,0,1)$, $B(b,0,0)$ and $C(0,c,0)$, $b, c$ are positive real numbers such that $b+c=1$. If $b,c$ change then there is a fixed sphere tangent to the plane $(ABC)$ and go through $D(1,1,1)$. Find the radius $R$ of the sphere.

Fortunately, take $E(1,1,0)$ we have $ED=d(E,(ABC))=1$. So $R=1$.

I'm looking for a geometric or analytic proof.

Answer 1

Since, there is a fixed sphere tangent to the plane that go through $D(1,1,1)$, so $R$ is half of distance between $(1,1,1)$ and the plane. The equation of plane is $\dfrac{x}{b}+\dfrac{y}{c}+z=1$ then $$R=\frac12\frac{\Big|\dfrac{1}{b}+\dfrac{1}{c}+1-1\Big|}{\sqrt{\dfrac{1}{b^2}+\dfrac{1}{c^2}+1}}$$