Câu hỏi 28

* Giả sử \(D\left( {a;b;c} \right).\,\,D \in AB \Rightarrow \overrightarrow {AD} //\overrightarrow {AB} \)

\(\left\{ \begin{array}{l}\overrightarrow {AD}  = \left( {a - 2;b - 1;c} \right)\\\overrightarrow {AB}  = \left( { - 1;1;2} \right)\end{array} \right. \Rightarrow \dfrac{{a - 2}}{{ - 1}} = \dfrac{{b - 1}}{1} = \dfrac{c}{2} \Leftrightarrow \left\{ \begin{array}{l}a + b - 3 = 0\,\,\,\,\left( 1 \right)\\2b - c - 2 = 0\,\,\left( 2 \right)\end{array} \right.\)

\(*\,\,\left\{ \begin{array}{l}\overrightarrow {CD}  = \left( {a - 1;b - 1;c} \right)\\\overrightarrow {{n_P}}  = \left( {1;1;1} \right)\end{array} \right.;\,\,\overrightarrow {CD} .\overrightarrow {{n_P}}  = 0 \Leftrightarrow a - 1 + b - 1 + c = 0 \Leftrightarrow a + b + c - 2 = 0\,\,\left( 3 \right)\)

* Giải hệ (1),(2),(3) \( \Rightarrow D\left( {\dfrac{5}{2};\dfrac{1}{2}; - 1} \right) \Rightarrow P = {x_D} + {y_D} + {z_D} = \dfrac{5}{2} + \dfrac{1}{2} - 1 = 2\).

Chọn B.