Đáp án:

- \( d \cap mp \, D = A \)  
- Lấy \( I \in d \), hạ \( IH \perp mp \, \alpha \)  
- \( mp \, \alpha \cap mp(P) = \Delta \)  
- Hạ \( HK \perp \Delta \)  
- \( \text{Góc } (\text{mp } P, \text{mp } \alpha) = \angle IKH \)

\( 0^\circ < \varphi \leq 90^\circ \)  \( \sin \varphi = \frac{IH}{IK} \geq \frac{IH}{IA} \)  

\( \varphi \) nhỏ nhất \( \Leftrightarrow \sin \varphi \) nhỏ nhất \( \Leftrightarrow K \equiv A \)  
\( \Leftrightarrow HI \perp \Delta  \Leftrightarrow \Delta \perp d \).  

\( \begin{cases} \vec{u}_\Delta \perp \vec{u}_d = (2, 1, 1) \\ \vec{u}_\Delta \perp \vec{n_P} = (1, 2, -1) \end{cases} \Rightarrow \vec{u}_\Delta = [\vec{u}_d, \vec{u}_{P}] = (-3, 3, 3) \parallel (-1, 1, 1). \)

\( \begin{cases} \vec{n}_\alpha \perp \vec{u_d} = (2, 1, 1) \\ \vec{n}_\alpha \perp \vec{u}_\Delta = (-1, 1, 1) \end{cases} \Rightarrow \vec{n}_\alpha = [\vec{u}_d, \vec{u}_\Delta] = (0, -3, 3) \parallel (0, 1, -1). \)

Mặt phẳng \( \alpha \) qua \( A( -1, -1, 3 ) \perp \vec{n}_\alpha = (0, 1, -1) \)  
\( \Rightarrow \text{pt mp } \alpha: y - z + 4 = 0 \Rightarrow \boxed{D} \)