Answer
$1v_1 + 0v_2 = \begin{bmatrix} 3 \\ 0 \\ 2 \end{bmatrix}$
$0v_1 + 1v_2 = \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}$
$1v_1 + 1v_2 = \begin{bmatrix} 1 \\ 0 \\ 5 \end{bmatrix}$
$1v_1 - 1v_2 = \begin{bmatrix} 5 \\ 0 \\ -1 \end{bmatrix}$
$1v_1 - 2v_2 = \begin{bmatrix} 7 \\ 0 \\ -4 \end{bmatrix}$
Work Step by Step
Given $v_1 = \begin{bmatrix} 3 \\ 0 \\ 2 \end{bmatrix}$ and $v_2 = \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}$, we can generate vectors in the span of $\{v_1, v_2\}$ by creating linear combinations:
$1v_1 + 0v_2 = v_1 = \begin{bmatrix} 3 \\ 0 \\ 2 \end{bmatrix}$
$0v_1 + 1v_2 = v_2 = \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}$
$1v_1 + 1v_2 = \begin{bmatrix} 1 \\ 0 \\ 5 \end{bmatrix}$
$1v_1 - 1v_2 = \begin{bmatrix} 5 \\ 0 \\ -1 \end{bmatrix}$
$1v_1 - 2v_2 = \begin{bmatrix} 7 \\ 0 \\ -4 \end{bmatrix}$
Of course, by selecting different weights, you can generate an infinite number of possible vectors.
