Phương pháp giải:

Lời giải chi tiết:

* Giả sử \(\overrightarrow {{a_d}} = \left( {a;b;c} \right)\,\,\left( {Dk:\,\,{a^2} + {b^2} + {c^2} \ne 0} \right)\).

* \(\left\{ \begin{array}{l}\overrightarrow {{a_d}} = \left( {a;b;c} \right)\\\overrightarrow {{n_Q}} = \left( {2; - 1; - 1} \right)\end{array} \right.;\,\,\overrightarrow {{n_Q}} \bot \overrightarrow {{a_d}} \Rightarrow \overrightarrow {{n_Q}} .\overrightarrow {{a_d}} = 0\)

\( \Leftrightarrow 2a - b - c = 0 \Leftrightarrow c = 2a - b \Rightarrow \overrightarrow {{a_d}} = \left( {a;b;2a - b} \right)\)

* \(\overrightarrow {{a_\Delta }} = \left( {1; - 2;2} \right)\). Ta có:

\(\cos \left( {\Delta ;d} \right) = \left| {\dfrac{{\overrightarrow {{a_d}} .\overrightarrow {{a_\Delta }} }}{{\left| {\overrightarrow {{a_d}} } \right|.\left| {\overrightarrow {{a_\Delta }} } \right|}}} \right| = \left| {\dfrac{{a - 2b + 2\left( {2a - b} \right)}}{{\sqrt {{a^2} + {b^2} + {{\left( {2a - b} \right)}^2}} \sqrt 9 }}} \right| = \cos \alpha \)

* Đặt \(t = \dfrac{a}{b}\,\,\left( {t \in \mathbb{R}} \right)\). Ta có:

\(\cos \alpha = \dfrac{1}{3}\dfrac{{\left| {5t - 4} \right|}}{{\sqrt {5{t^2} - 4t + 2} }} = \dfrac{1}{3}\sqrt {\dfrac{{{{\left( {5t - 4} \right)}^2}}}{{5{t^2} - 4t + 2}}} \).

Đặt \(f\left( t \right) = \dfrac{{{{\left( {5t - 4} \right)}^2}}}{{5{t^2} - 4t + 2}} \Rightarrow f'\left( t \right) = \dfrac{{2\left( {5t - 4} \right)\left( {10t + 2} \right)}}{{{{\left( {5{t^2} - 4t + 2} \right)}^2}}}\)

\(f'\left( t \right) = 0 \Leftrightarrow \left[ \begin{array}{l}t = \dfrac{4}{5}\\t = - \dfrac{1}{5}\end{array} \right.\)

* \(f\left( {\dfrac{4}{5}} \right) = 0,\,\,f\left( { - \dfrac{1}{5}} \right) = \dfrac{{25}}{3}\). Hàm số \(\cos \alpha \) nghịch biến trên \(\left[ {{0^0};{{90}^0}} \right]\) \( \Rightarrow f{\left( t \right)_{\min }} \Leftrightarrow t = - \dfrac{1}{5}\).

Chọn \(a = 1 \Rightarrow b = - 5 \Rightarrow c = 7\).

Chọn A.