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 \( \langle x, y, z\rangle=\langle \)

To find the vector equation of the line through \( (0,0,0) \) that is parallel to the line given by \( \mathbf{r}=\langle 3-2 t, 9+5 t, 2+3 t\rangle \), we first identify the direction vector of the given line. The coefficients of \( t \) in the vector equation imply the direction vector is \( \langle -2, 5, 3 \rangle \). Using the point \( (0,0,0) \) as the starting point, the vector equation for the new line is: \[ \langle x, y, z\rangle=\langle 0, 0, 0\rangle + t\langle -2, 5, 3\rangle = \langle -2t, 5t, 3t\rangle. \] Next, to find the parametric equations, we can write: \[ x = -2t, \quad y = 5t, \quad z = 3t. \] And there you have both the vector equation and the parametric equations of the line!