$1)2x^2+2y^2+2z^2-8x+8y-4z=0\\ \Leftrightarrow x^2+y^2+z^2-4x+4y-2z=0\\ \Leftrightarrow (x-2)^2+(y+2)^2+(z-1)^2=9$
Tâm $I(2;-2;1),R=3$
$2)$Tâm là trung điểm $AB:I(3;2;1)$
Bán kính $IA=\sqrt{(3-1)^2+(2-2)^2+(1-3)^2}=2\sqrt{2}$
$\Rightarrow (O):(x-3)^2+(y-2)^2+(z-1)^2=8$
$3)$Gọi tâm mặt cầu là $I(a,b,c)$
$IA=IB=IC=ID\\ \Rightarrow \left\{\begin{array}{l} IA^2=IB^2\\ IA^2=IC^2 \\ IA^2=ID^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} (a+1)^2+(b+1)^2+(c+1)^2=(a-1)^2+b^2+c^2\\ (a+1)^2+(b+1)^2+(c+1)^2=a^2+(b-2)^2+c^2 \\ (a+1)^2+(b+1)^2+(c+1)^2=a^2+b^2+(c-3)^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 4a+2b+2c+2=0\\ 2a+6b+2c-1=0 \\ 2a+2b+8c-6=0\end{array} \right.\\\Leftrightarrow \left\{\begin{array}{l} a=-\dfrac{37}{34}\\ b=\dfrac{7}{34} \\ c=\dfrac{33}{34}\end{array} \right.\\ \Rightarrow IA^2=\dfrac{6179}{1156}\\ \Rightarrow (O):\left(x+\dfrac{37}{34}\right)^2+\left(y-\dfrac{7}{34} \right)^2+\left(z-\dfrac{33}{34}\right)^2=\dfrac{6179}{1156}$
