Lời giải:

\(N(a, b, c)\)  
\( \begin{cases} \overrightarrow{MN} = (a - 1, b + 1, c - 1) // \vec{n}_P = (1, -2, -3) \\I \left(\frac{a + 1}{2}, \frac{b - 1}{2}, \frac{c + 1}{2}\right) \in \text{mp} P: x - 2y - 3z + 14 = 0 \end{cases}\)  

\(\Leftrightarrow \begin{cases} \frac{a-1}{1}= \frac{b+1}{-2} = \frac{c-1}{-3} \\ \frac{a + 1}{2} - 2\left(\frac{b - 1}{2}\right) - 3\left(\frac{c + 1}{2}\right) + 14 = 0 \end{cases}\)

\(\Leftrightarrow \begin{cases}
-2a - b = -1, \\
-3a - c = -4, \\
\frac{a}{2} - b - \frac{3c}{2} = -14.
\end{cases} \Leftrightarrow \begin{cases} a = -1\\ b = 3 \\ c = 7 \end{cases}\)

\(\Leftrightarrow N(-1, 3, 7) \Rightarrow \boxed{D}\).