Answer
The sphere has center $(\frac{25}{3}, 1, -\frac{11}{3})$ and radius $\frac{2\sqrt {83}}{3}$ (See process for proof).
Work Step by Step
First, we give $P$ the general coordinates $(x, y, z)$.
Then, using the formula for distance, the distance from $P$ to $A(-1, 5, 3)$ becomes:
$|PA| =\sqrt {(x+1)^{2} + (y-5)^{2} + (z-3)^{2}}$
And the distance from $P$ to $B(6, 2, -2)$ becomes:
$|PB| =\sqrt {(x-6)^{2} + (y-2)^{2} + (z+2)^{2}}$.
The problem says that $|PA|$ is twice $|PB|$, and so $|PA| = 2|PB|$. Thus we have that:
$\sqrt {(x+1)^{2} + (y-5)^{2} + (z-3)^{2}} = 2\sqrt {(x-6)^{2} + (y-2)^{2} + (z+2)^{2}}$
Squaring both sides we get:
$(x+1)^{2} + (y-5)^{2} + (z-3)^{2} = 4((x-6)^{2} + (y-2)^{2} + (z+2)^{2})$
Expanding and distributing, we get:
$x^{2} +2x+1 + y^{2}-10y+25 + z^{2}-6z+9 = 4x^{2}-48x+144 + 4y^{2}-16y+16 + 4z^{2}+16z+16$
Now we combine like terms and leave the constant term on the right side:
$3x^{2}-50x + 3y^{2}-6y+ 3z^{2}+22z= -141$
To complete the square more easily, we divide both sides by 3:
$x^{2}-\frac{50}{3}x + y^{2}-2y+ z^{2}+\frac{22}{3}z= -47$
Now we complete the square for each variable:
$x^{2}-\frac{50}{3}x + \frac{625}{9} + y^{2}-2y+1+ z^{2}+\frac{22}{3}z+\frac{121}{9}= -47 + \frac{625}{9} +1 + \frac{121}{9}$
$(x-\frac{25}{3})^{2} + (y-1)^{2}+ (z+\frac{11}{3})^{2}= \frac{332}{9}$
This is the equation of a sphere with center $(\frac{25}{3}, 1, -\frac{11}{3})$ and radius $\frac{\sqrt {332}}{3}$ (or $\frac{2\sqrt {83}}{3}$).
