Đáp án:
a) Tâm mặt cầu I là trung điểm của AB
$ \Rightarrow I\left( {\dfrac{1}{2};4;2} \right)$
$\begin{array}{l}
\Rightarrow \left( C \right):{\left( {x - \dfrac{1}{2}} \right)^2} + {\left( {y - 4} \right)^2} + {\left( {z - 2} \right)^2} = I{A^2}\\
\Rightarrow \left( C \right){\left( {x - \dfrac{1}{2}} \right)^2} + {\left( {y - 4} \right)^2} + {\left( {z - 2} \right)^2} = \dfrac{{97}}{4}\\
b)I \in Oy\\
\Rightarrow I\left( {0;y;0} \right)\\
\Rightarrow I{A^2} = I{B^2}\\
\Rightarrow {2^2} + {\left( {y - 1} \right)^2} + {5^2} = {3^2} + {\left( {y - 7} \right)^2} + {1^2}\\
\Rightarrow 4 + {y^2} - 2y + 1 + 25\\
= 9 + {y^2} - 14y + 49 + 1\\
\Rightarrow 12y = 29\\
\Rightarrow y = \dfrac{{29}}{{12}}\\
\Rightarrow I{A^2} = 29 + \dfrac{{{{17}^2}}}{{144}} = \dfrac{{4465}}{{144}}\\
\Rightarrow \left( C \right):{x^2} + {\left( {y - \dfrac{{29}}{{12}}} \right)^2} + {z^2} = \dfrac{{4465}}{{144}}
\end{array}$
