Answer
A line that passes through $(0, -3, 1)$ and is parallel to $2\mathbf{i}+3\mathbf{k}$
Work Step by Step
Step 1
From a vector-valued function: \[ \mathbf{r} = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \] the parametric equations can be obtained: \[ \begin{align*} x &= x(t) \\ y &= y(t) \\ z &= z(t) \end{align*} \] The graph can be identified by remembering certain forms, using trigonometric relationships, converting to Cartesian form, or by plotting some points and drawing a smooth curve through them.
Step 2
From: \[ \mathbf{r} = 2t\mathbf{i} - 3\mathbf{j} + (1+3t)\mathbf{k} \] we have parametric equations: \[ \begin{align*} x &= 2t \\ y &= -3 \\ z &= 1+3t \end{align*} \] This has the form of a parametric representation of a line: \[ \begin{align*} x &= x_0 + at \\ y &= y_0 + bt \\ z &= z_0 + ct \end{align*} \] where $(x_0, y_0, z_0)$ is a point on the line, and it is parallel to the vector $\langle a, b, c \rangle$. So, here this line passes through $(0, -3, 1)$ and is parallel to $\langle 2, 0, 3 \rangle$.
Result: A line that passes through $(0, -3, 1)$ and is parallel to $2\mathbf{i}+3\mathbf{k}$.
