Tứ diện \(ABCD\) có \(A\left( {3;3;0} \right),\,\,...

A

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

* Giả sử tâm \(I\left( {a;b;c} \right)\)

\(\begin{array}{l}\left\{ \begin{array}{l}IA = IB\\IB = IC\\IC = ID\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}I{A^2} = I{B^2}\\I{B^2} = I{C^2}\\I{C^2} = I{D^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\left( {a - 3} \right)^2} + {\left( {b - 3} \right)^2} + {c^2} = {a^2} + {\left( {b - 3} \right)^2} + {\left( {c - 3} \right)^2}\\{a^2} + {\left( {b - 3} \right)^2} + {\left( {c - 3} \right)^2} = {\left( {a - 3} \right)^2} + {b^2} + {\left( {c - 3} \right)^2}\\{\left( {a - 3} \right)^2} + {b^2} + {\left( {c - 3} \right)^2} = {\left( {a - 3} \right)^2} + {\left( {b - 3} \right)^2} + {\left( {c - 3} \right)^2}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} - 6a + 9 =  - 6c + 9\\ - 6b + 9 =  - 6a + 9\\0 =  - 6b + 9\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = c\\b = a\\b = \dfrac{3}{2}\end{array} \right. \Rightarrow a = b = c = \dfrac{3}{2} \Rightarrow I\left( {\dfrac{3}{2};\dfrac{3}{2};\dfrac{3}{2}} \right)\end{array}\)

Chọn A.