Trong không gian với hệ tọa độ Oxyz, cho điểm H(1;2;- 2).

Gọi \(A\left( {a;0;0} \right),B\left( {0;b;0} \right),C\left( {0;0;c} \right)\) lần lượt thuộc các trục tọa độ Ox, Oy, Oz.

Khi đó ta có phương trình \(\left( \alpha  \right)\) đi qua các điểm A, B, C: \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\) 

\(H \in \left( \alpha  \right) \Rightarrow \frac{1}{a} + \frac{2}{b} - \frac{2}{c} = 1\,\,\,\left( 1 \right)\)

Theo đề bài ta có H là trực tâm \(\Delta ABC \Rightarrow \left\{ \begin{array}{l}
\overrightarrow {AH}  \bot \overrightarrow {BC} \\
\overrightarrow {BH}  \bot \overrightarrow {AC} 
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\overrightarrow {AH} .\overrightarrow {BC}  = 0\\
\overrightarrow {BH} .\overrightarrow {AC}  = 0
\end{array} \right.\) 

Ta có: \(\left\{ \begin{array}{l}
\overrightarrow {AH}  = \left( {1 - a;2; - 2} \right),\overrightarrow {BC}  = \left( {0; - b;c} \right)\\
\overrightarrow {BH}  = \left( {1;2 - b; - 2} \right),\overrightarrow {AC}  = \left( { - a;0;c} \right)
\end{array} \right.\) 

\(\begin{array}{l}
\left\{ \begin{array}{l}
\overrightarrow {AH} .\overrightarrow {BC}  = 0\\
\overrightarrow {BH} .\overrightarrow {AC}  = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
 - 2b - 2c = 0\\
 - a - 2c = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a =  - 2c\\
b =  - c
\end{array} \right.\\
 \Rightarrow \left( 1 \right) \Leftrightarrow \frac{1}{{ - 2c}} + \frac{2}{{ - c}} - \frac{2}{c} = 1 \Rightarrow  - \frac{9}{{2c}} = 1 \Leftrightarrow c =  - \frac{9}{2}
\end{array}\)

\( \Rightarrow \left\{ \begin{array}{l}
a =  - 2c = 9\\
b =  - c = \frac{9}{2}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
A\left( {9;0;0} \right)\\
B\left( {0;\frac{9}{2}0} \right)\\
C\left( {0;0; - \frac{9}{2}} \right)
\end{array} \right.\) 

Gọi \(I\left( {{x_0};{y_0};{z_0}} \right)\) là tâm mặt cầu ngoại tiếp chóp tứ giác OABC.

\(\begin{array}{l}
 \Rightarrow \left\{ \begin{array}{l}
OI = IA\\
OI = IB\\
OI = IC
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x_0^2 + y_0^2 + z_0^2 = {\left( {{x_0} - 9} \right)^2} + y_0^2 + z_0^2\\
x_0^2 + y_0^2 + z_0^2 = x_0^2 + {\left( {{y_0} - \frac{9}{2}} \right)^2} + z_0^2\\
x_0^2 + y_0^2 + z_0^2 = x_0^2 + y_0^2 + {\left( {{z_0} + \frac{9}{2}} \right)^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x_0^2 = {\left( {{x_0} - 9} \right)^2}\\
y_0^2 = {\left( {{y_0} - \frac{9}{2}} \right)^2}\\
z_0^2 = {\left( {{z_0} + \frac{9}{2}} \right)^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x_0} =  - {x_0} + 9\\
{y_0} =  - {y_0} + \frac{9}{2}\\
{z_0} =  - {z_0} - \frac{9}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x_0} = \frac{9}{2}\\
{y_0} = \frac{9}{4}\\
{z_0} =  - \frac{9}{4}
\end{array} \right. \Rightarrow I\left( {\frac{9}{2};\frac{9}{4};\frac{9}{4}} \right) \Rightarrow R = OI = \frac{{9\sqrt 6 }}{4}\\
 \Rightarrow {S_{\left( I \right)}} = 4\pi {R^2} = 4\pi .{\left( {\frac{{9\sqrt 6 }}{4}} \right)^2} = \frac{{243\pi }}{2}
\end{array}\)