Answer
Sphere centered at $C\left(-\frac{1}{3}, 2, \frac{4}{3}\right)$ of radius $r=\frac{2 \sqrt{11}}{3}$
Work Step by Step
It's given that
\[
2 d(P, B)=d(P, A)
\]
where $A, B, P$ have coordinates
\[
P(x, y, z), \quad B(0,1,1) \text { and } A(1,-2,0)
\]
and then
\[
\begin{aligned}
4 d(P, B) = d^{2}(P, A) \\
(2+y)^{2}+(-1+x)^{2}+(z-0)^{2}=4\left[(-1+z)^{2}+(x-0)^{2}+(-1+y)^{2}\right] \\
\Rightarrow y^{2}+4 y+4+z^{2}+x^{2}-2 x+1=& 4 y^{2}-8 y+4+4 z^{2}-8 z+4+4 x^{2} \\
\Rightarrow 3 z^{2}+2 x-12 y+3 x^{2}+3 y^{2}-8 z=& \\
\Rightarrow y^{2}-4 y+x^{2}+\frac{2}{3} x+z^{2}-\frac{8}{3} z=-1 & \\
\Rightarrow\left(\frac{2}{3} x+x^{2}+\frac{1}{9}\right)+\left(y^{2}-4 y+4\right)+\left(z^{2}-\frac{8}{3} z+\frac{16}{9}\right)=\frac{1}{9}+4+\frac{16}{9}-1 & \\
\Rightarrow\left(\frac{1}{3}+x\right)^{2}+(-2+y)^{2}+(-4 / 3+z)^{2}=\frac{44}{9}=\left(\frac{2 \sqrt{11}}{3}\right)^{2}
\end{aligned}
\]
This equation corresponds to a sphere centered at $C\left(-\frac{1}{3}, 2, \frac{4}{3}\right)$ with the radius
\[
\frac{2 \sqrt{11}}{3}=r
\]