Trong không gian \(Oxyz\), cho bốn điểm \(A( {a;0;0} ,B( {0;b;0} ,C( {0;0;c} D( {1;2; - 1}

Phương trình mặt phẳng \(\left( {ABC} \right):\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1{\rm{ }}\left( {acb \ne 0} \right)\).

Vì \[D\left( {1;2; - 1} \right)\]thuộc mặt phẳng nên ta có: \[\frac{1}{a} + \frac{2}{b} - \frac{1}{c} = 1\].

Ta có \(d\left( {O,\left( {ABC} \right)} \right) = \frac{1}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} }}\)

Ta có \({\left( {\frac{1}{a} + \frac{2}{b} - \frac{1}{c}} \right)^2} = {\left( {\frac{1}{a}{\rm{ + 2}}{\rm{.}}\frac{1}{b}{\rm{ + }}\left( { - 1} \right){\rm{.}}\frac{1}{c}} \right)^2} \le 6\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} \right){\rm{ }}\)

\( \Rightarrow 1 \le \sqrt 6 .\sqrt {\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} \right){\rm{ }}} \Rightarrow \frac{1}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}{\rm{ }}} }} \le \sqrt 6 \).

Suy ra \(d\left( {O,\left( {ABC} \right)} \right) \le \sqrt 6 \)

\({d_{\max }} = \sqrt 6 \Leftrightarrow \left\{ \begin{array}{l}a = 2b = - c\\\frac{1}{a} + \frac{2}{b} - \frac{1}{c} = 1{\rm{ }}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 6\\b = 3\\c = - 6\end{array} \right. \Rightarrow a + b + c = 3\).