Đáp án:

\( B(0, -2, 1) \in d_1: \text{ Mp}(A, d_1) \text{ có vectơ pháp: } \vec{n}_1 = \left[\begin{split} \vec{u}_{d_1}=(2, 1, 2) \\ \overrightarrow{AB} = (0, -1, -1) \end{split} \right] = (1, 2, -2). \)  

\( C(-1, 3, 1) \in d_2: \text{ Mp}(A, d_2) \text{ có vectơ pháp: } \vec{n}_2 = \left[\begin{split} \vec{u}_{d_2}=(1, 1, 2) \\ \overrightarrow{AC} = (-1, 4, -1) \end{split} \right] = (-9, -1, 5). \)  

 \(  \vec{u}_\Delta = \left[\begin{split} \vec{n}_1=(1, 2, -2) \\ \vec{n}_2=(-9, -1, 5) \end{split} \right]  = (8, 13, 17). \)

Chọn \( \boxed{D}. \)

Cách 2:
\( \Delta = \text{mp}(A, d_1) \cap \text{mp}(A, d_2) \).  

Mp \( (A, d_1) \): Qua \( A(0, -1, 2), B(0, -2, 1), C(2, -1, 3) \).  
\( \Rightarrow \text{mp}(A, d_1): -x - 2y + 2z - 6 = 0. \)  

Mp \( (A, d_2) \):  Qua \( A(0, -1, 2), D(-1, 3, 1), E(0, 4, 3) \).  
\( \Rightarrow \text{mp}(A, d_2): -9x - y + 5z - 11 = 0. \)  

\( \vec{u}_\Delta = \left[\begin{split} \vec{n}_1 =(-1, -2, 2) \\ \vec{n}_2 = (-9, -1, 5)\end{split} \right]  = (8, 13, 17). \)  

\( \vec{u}_\Delta = (8, 13, 17). \)