Câu hỏi 25

\(A \in tia\,\,Ox \Rightarrow A\left( {a;0;0} \right)\,\,\left( {a > 0} \right)\).

\(\begin{array}{l}\left\{ \begin{array}{l}\overrightarrow {BA}  = \left( {a - 3;1; - 4} \right)\\\overrightarrow {BC}  = \left( { - 2;4; - 5} \right)\end{array} \right. \Rightarrow \overrightarrow n  = \left[ {\overrightarrow {BA} ;\overrightarrow {BC} } \right] = \left( {11;5a - 7;4a - 10} \right)\\{S_{\Delta ABC}} = \dfrac{1}{2}\left| {\overrightarrow n } \right| = \dfrac{1}{2}\sqrt {121 + {{\left( {5a - 7} \right)}^2} + {{\left( {4a - 10} \right)}^2}}  = \dfrac{{\sqrt {134} }}{2}\\ \Rightarrow 121 + {\left( {5a - 7} \right)^2} + {\left( {4a - 10} \right)^2} = 134\\ \Leftrightarrow 41{a^2} - 150a + 136 = 0 \Leftrightarrow \left[ \begin{array}{l}a = 2\,\,\left( {tm} \right)\\a = \dfrac{{68}}{{41}}\,\,\left( {ktm} \right)\end{array} \right. \Rightarrow A\left( {2;0;0} \right)\end{array}\)

Chọn B.