The lines 

​$$\frac{x}{1} = \frac{y}{2} = \frac{z}{3} \quad \text{and} \quad \frac{x - 3}{-3} = \frac{y - 4}{-6} = \frac{z - 5}{-9} \quad \text{are:}$$​​

Correct option is D

​$$\begin{aligned}&\textbf{1. Line 1:} \\&\quad \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \\&\quad \textbf{Direction vector } (\vec{d_1}): \vec{d_1} = (1, 2, 3) \\&\quad \textbf{Point on the line } (P_1): (0, 0, 0) \\\\&\textbf{2. Line 2:} \\&\quad \frac{x - 3}{-3} = \frac{y - 4}{-6} = \frac{z - 5}{-9} \\&\quad \textbf{Direction vector } (\vec{d_2}): \vec{d_2} = (-3, -6, -9) \\&\quad \textbf{Point on the line } (P_2): (3, 4, 5) \\\\&\textbf{Check if the Direction Vectors are Scalar Multiples} \\&\quad \vec{d_2} = -3 \cdot \vec{d_1} \\&\quad (-3, -6, -9) = -3 \cdot (1, 2, 3) \\\\&\text{Since } \vec{d_2} \text{ is a scalar multiple of } \vec{d_1}, \text{ the lines are } \textbf{parallel}.\end{aligned}$$​​