Giả sử \(A\left( a;\,0;\,0 \right),\,B\left( 0;\,b;\,0 \right),\,C\left( 0;\,0;\,c \right)\)

Do \(A,\,B,\,C\) nằm trên các tia \(Ox,\,Oy,\,Oz\) nên \(a,\,b,\,c\,>\,0\).

\(O{{A}^{2}}\,+\,O{{B}^{2}}\,+\,O{{C}^{2}}\,=\,27\,\Leftrightarrow \,{{a}^{2}}\,+\,{{b}^{2}}\,+\,{{c}^{2}}\,=\,27\)

Ta có \(\left( \alpha  \right):\,\frac{x}{a}\,+\,\frac{y}{b}\,+\,\frac{z}{c}\,=\,1\,\Leftrightarrow \,bcx\,+\,cay\,+\,abz\,-\,abc\,=\,0\)

Mặt cầu \(\left( S \right):\,{{x}^{2}}\,+\,{{y}^{2}}\,+\,{{z}^{2}}\,=\,3\) có tâm O và bán kính \(R\,=\,\sqrt{3}\)

Do \(\left( \alpha  \right)$ tiếp xúc với \(\left( S \right)\) nên \(d\left( O;\,\left( \alpha  \right) \right)\,=\,\sqrt{3}\,\Leftrightarrow \,\frac{abc}{\sqrt{{{a}^{2}}{{b}^{2}}\,+\,{{b}^{2}}{{c}^{2}}\,+\,{{c}^{2}}{{a}^{2}}}}\,=\,\sqrt{3}\)

\(\Leftrightarrow \,{{a}^{2}}{{b}^{2}}{{c}^{2}}\,=\,3\left( {{a}^{2}}{{b}^{2}}\,+\,{{b}^{2}}{{c}^{2}}\,+\,{{c}^{2}}{{a}^{2}} \right)\,\Leftrightarrow \,\frac{1}{{{a}^{2}}}\,+\,\frac{1}{{{b}^{2}}}\,+\,\frac{1}{{{c}^{2}}}\,=\,\frac{1}{3}\)

Ta có \(\left( {{a}^{2}}\,+\,{{b}^{2}}\,+\,{{c}^{2}} \right)\left( \frac{1}{{{a}^{2}}}\,+\,\frac{1}{{{b}^{2}}}\,+\,\frac{1}{{{c}^{2}}} \right)\,\ge \,3.\sqrt[3]{{{a}^{2}}{{b}^{2}}{{c}^{2}}}.\frac{3}{\sqrt[3]{{{a}^{2}}{{b}^{2}}{{c}^{2}}}}\,=\,9\)

Mà theo giả thiết \(\left( {{a}^{2}}\,+\,{{b}^{2}}\,+\,{{c}^{2}} \right)\left( \frac{1}{{{a}^{2}}}\,+\,\frac{1}{{{b}^{2}}}\,+\,\frac{1}{{{c}^{2}}} \right)\,=\,9\) nên từ đó ta có \(a\,=\,b\,=\,c\,=\,3\)

\({V_{OABC}} = \frac{{abc}}{6} = \frac{9}{2} \Rightarrow {S_{\Delta ABC}} = \frac{{3{V_{OABC}}}}{{d\left( {O,\left( \alpha  \right)} \right)}} = \frac{{27}}{{2\sqrt 3 }} = \frac{{9\sqrt 3 }}{2}\)