Đáp án:

Cách 1: Thử !!!  
\( \vec{n}_Q = (1,-2,-2) \)  
\( M(1,3,2) \).  
Chọn \( N \) sao cho \( \overrightarrow{MN} = \vec{n}_Q  \Rightarrow N(2,1,0) \)  

Mp \( P: ax + by + cz + d = 0 \).  

Vì \( d(O,(P)) = 1 \Rightarrow d \neq 0 \). Chọn \( d = 1 \).  

Mp \( P: ax + by + cz + 1 = 0 \).  

- Mp \( P \) qua \( M(1,3,2) \Rightarrow a + 3b + 2c + 1 = 0 \)  

- Mp \( P \) qua \( N(2,1,0) \Rightarrow 2a + b + 1 = 0 \)  

- \( d(O, (P)) = 1 \Rightarrow \frac{1}{\sqrt{a^2 + b^2 + c^2}} = 1 \)  

Tóm lại:  
\( \begin{cases} a + 3b + 2c + 1 = 0 \\ 2a + b + 1 = 0 \\ a^2 + b^2 + c^2 = 1 \end{cases} \Leftrightarrow \begin{cases} b = -1 - 2a \\ c = \frac{1}{2}(-1 - a - 3b) = \frac{1}{2}(5a + 2) \\ a^2 + (-1 - 2a)^2 +  \frac{1}{4} (5a + 2)^2 - 1 = 0 \end{cases}\) (*)  

(*) \(\Rightarrow 45a^2 + 36a + 4 = 0 \Rightarrow \begin{cases} a = -\frac{2}{3}, b = \frac{1}{3}, c = -\frac{2}{3} \\ a = -\frac{2}{15}, b = -\frac{11}{15}, c = \frac{10}{15} \end{cases}\)

Mp \( P: \left[\begin{split} &2x - y + 2z - 3 = 0 \\ &2x + 11y - 10z - 15 = 0 \end{split} \right. \Rightarrow \boxed{C}\)