Answer
the center is $($ $\frac{25}{3}$ $,$ $1$ $,$ $-\frac{11}{3}$ $)$
the radius is $\sqrt \frac{332}{9}$
Work Step by Step
As we know, the points P to A $($ $-1$, $5$, $3$ $)$ is twice the distance from P to B $($ $6$, $2$, $-2$ $)$,
$|$ $\overrightarrow{PA}$ $|$ $=$ $\sqrt{(x+1)^{2}+(y-5)^{2}+(z-3)^{2}}$
$|$ $\overrightarrow{PB}$ $|$ $=$ $\sqrt{(x-6)^{2}+(y-2)^{2}+(z+2)^{2}}$
$|$ $\overrightarrow{PA}$ $|$ $=$ $2$ $|$ $\overrightarrow{PB}$ $|$
Square both sides,
$(x+1)^{2}$ $+$ $(y-5)^{2}$ $+$ $(z-3)^{2}$ $=$ $4$ $[ $$(x-6)^{2}$ $+$ $(y-2)^{2}$ $+$ $(z+2)^{2}$ $]$
$x^{2}$ $+$ $2x$ $+$ $1$ $+$ $y^{2}$ $-$ $10y$ $+$ $25$ $+$ $z^{2}$ $-$ $6z$ $+$ $9$ $=$ $4$ $($ $x^{2}$ $-$ $12x$ $+$ $36$ $+$ $y^{2}$ $-$ $4y$ $+$ $4$ $+$ $z^{2}$ $+$ $4z$ $+$ $4$ $)$
$-141$ $=$ $3x^{2}$ $-$ $50x$ $+$ $3y^{2}$ $-$ $6y$ $+$ $3z^{2}$ $+$ $22z$
$-47$ $=$ $x^{2}$ $-$ $\frac{56}{3}x$ $+$ $y^{2}$ $-$ $2y$ $+$ $z^{2}$ $+$ $\frac{22}{3}z$ $+$ $(\frac{56}{6})^{2}$ $+$ $1^{2}$ $+$ $(\frac{22}{6})^{2}$
$(x-\frac{25}{3})^{2}$ $+$ $(y-1)^{2}$ $+$ $(z+\frac{11}{3})^{2}$ $=$ $\frac{332}{9}$
so, the center is $($ $\frac{25}{3}$ $,$ $1$ $,$ $-\frac{11}{3}$ $)$
the radius is $\sqrt \frac{332}{9}$
