Find both the vector equation and the parametric equations of the line through \( (0,0,0) \) that is parallel to the line \( \mathbf{r}=\langle 3-2 t, 9+5 t, 2+3 t\rangle \), where \( t=0 \) corresponds to the given point. The vector equation is \( \langle x, y, z\rangle=\langle\square, \square+\square \)

To find the vector equation of the line, we start by identifying the direction vector of the given line. From the line equation \( \mathbf{r}=\langle 3-2t, 9+5t, 2+3t \rangle \), we can extract the coefficients of \( t \) to determine the direction vector, which is \( \langle -2, 5, 3 \rangle \). Since the line must pass through the point \( (0,0,0) \) and be parallel to the given line, the vector equation is: \[ \langle x, y, z \rangle = \langle 0, 0, 0 \rangle + t \langle -2, 5, 3 \rangle = \langle -2t, 5t, 3t \rangle. \] Thus, the completed vector equation is: \[ \langle x, y, z \rangle = \langle -2t, 5t, 3t \rangle. \] Next, we can derive the parametric equations by expressing each component in terms of \( t \): - \( x = -2t \) - \( y = 5t \) - \( z = 3t \) These parametric equations define the line parallel to the initial line and passing through the origin.