91Ó°ÊÓ

To find the direction vector \(\vec{d}\) of the line passing through the two points, subtract the coordinates of the starting point from the coordinates of the ending point: $$\vec{d} = (1 - 0, 2 - 0, 3 - 0) = (1, 2, 3).$$

Using the direction vector, we can write the parametric equations of the line as follows: $$\begin{cases} x = x_0 + td_x \\ y = y_0 + td_y \\ z = z_0 + td_z \end{cases}$$ where \((x_0, y_0, z_0)\) are the coordinates of the starting point \((0,0,0)\), \((d_x, d_y, d_z)\) are the components of the direction vector \((1,2,3)\), and \(t\) is the parameter that ranges from \(0\) to \(1\) since it's a line segment. Plugging in the values, we get: $$\begin{cases} x = 0 + t(1) \\ y = 0 + t(2) \\ z = 0 + t(3) \end{cases}$$ The parametric equations of the line segment joining the points \((0,0,0)\) and \((1,2,3)\) are: $$\begin{cases} x = t \\ y = 2t \\ z = 3t \end{cases}$$ with the parameter \(t\) ranging from \(0\) to \(1\).