Đáp án đúng là: D

Gọi \[A\left( {a;0;0} \right),B\left( {0;b;0} \right),C\left( {0;0;c} \right)\].

Tam giác \[ABC\] có trọng tâm \[G\left( { - 6; - 12;18} \right)\] nên ta có:

\[\left\{ \begin{array}{l}\frac{{a + 0 + 0}}{3} = - 6\\\frac{{0 + b + 0}}{3} = - 12\\\frac{{0 + 0 + c}}{3} = 18\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = - 18\\b = - 36\\c = 54\end{array} \right.\].

Suy ra \[A\left( { - 18;0;0} \right),B\left( {0; - 36;0} \right),C\left( {0;0;54} \right)\].

Gọi \[I\left( {x;y;z} \right)\], ta có: \[\left\{ \begin{array}{l}IO = IA\\IA = IB\\IB = IC\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}I{O^2} = I{A^2}\\I{A^2} = I{B^2}\\I{B^2} = I{C^2}\end{array} \right.\].

\[ \Leftrightarrow \left\{ \begin{array}{l}{x^2} + {y^2} + {z^2} = {\left( {x + 18} \right)^2} + {y^2} + {z^2}\\{\left( {x + 18} \right)^2} + {y^2} + {z^2} = {x^2} + {\left( {y + 36} \right)^2} + {z^2}\\{x^2} + {\left( {y + 36} \right)^2} + {z^2} = {x^2} + {y^2} + {\left( {z - 54} \right)^2}\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}36x + 324 = 0\\36x + 324 - 72y - 1296 = 0\\72y + 1296 + 108z - 2916 = 0\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}x = - 9\\y = - 18\\z = 27\end{array} \right.\].

Vậy tâm của mặt cầu là \[I\left( { - 9; - 18;27} \right).\]