1. **State the problem:** Find the vector equation of a line parallel to the line given by $$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ -3 \end{pmatrix} + t \begin{pmatrix} -2 \\ 5 \\ 2 \end{pmatrix}$$ that passes through the point $(2,5,-3)$. 2. **Recall the formula for a vector equation of a line:** $$\vec{r} = \vec{r_0} + t \vec{d}$$ where $\vec{r_0}$ is a point on the line and $\vec{d}$ is the direction vector. 3. **Identify the direction vector:** The given line has direction vector $$\vec{d} = \begin{pmatrix} -2 \\ 5 \\ 2 \end{pmatrix}$$ Since the new line is parallel, it has the same direction vector. 4. **Use the given point as $\vec{r_0}$:** $$\vec{r_0} = \begin{pmatrix} 2 \\ 5 \\ -3 \end{pmatrix}$$ 5. **Write the vector equation of the new line:** $$\vec{r} = \begin{pmatrix} 2 \\ 5 \\ -3 \end{pmatrix} + t \begin{pmatrix} -2 \\ 5 \\ 2 \end{pmatrix}$$ **Final answer:** $$\boxed{\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ 5 \\ -3 \end{pmatrix} + t \begin{pmatrix} -2 \\ 5 \\ 2 \end{pmatrix}}$$