Equivalent parametric solution sets I have this question that I've never seen before, I've only ever learned about how to find the parametric version of a solution set, but I've never learned how to change it in any way...
I will type out the question as well as include the image

A system of linear equations has the following solution set (in parametric form):
$$\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix} + t\begin{bmatrix}
1 \\
0 \\
-1
\end{bmatrix} + s\begin{bmatrix}
0 \\
1 \\
-1
\end{bmatrix}$$

Which of the following is also a description of this solution set? Select all that apply.
A. $\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} + t\begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix} + s\begin{bmatrix}
0 \\
-2 \\
2
\end{bmatrix}$
B. $\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix} + t\begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix} + s\begin{bmatrix}
2 \\
-2 \\
0
\end{bmatrix}$
C. $\begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix} + t\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix} + s\begin{bmatrix}
1 \\
0 \\
-1
\end{bmatrix}$
D. $\begin{bmatrix}
-1 \\
2 \\
0
\end{bmatrix} + t\begin{bmatrix}
1 \\
0 \\
-1
\end{bmatrix} + s\begin{bmatrix}
3 \\
-3 \\
0
\end{bmatrix}$
E. $\begin{bmatrix}
2 \\
0 \\
0
\end{bmatrix} + t\begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix} + s\begin{bmatrix}
0 \\
1 \\
-1
\end{bmatrix}$
image of question
The answer key in my book says that the answer is A) and D), but no explanation. Any help will be very appreciated!

Two sets $\{V + tW_1 + sW_2\}$, $\{\widetilde V + t\widetilde W_1 + s\widetilde W_2\}$ representing planes in $\mathbb{R}^3$ are the same if and only if both of the following conditions hold:

*

*$N = W_1 \times W_2$ and $\widetilde N = \widetilde W_1 \times \widetilde W_2$ are the same up to scaling (where $\times$ is the cross product)

*$N \cdot V = N \cdot \widetilde V$ (where $\cdot$ is the dot product).

For example, for the solution set in your question, $N = \begin{bmatrix} 
1 \\ 0 \\ -1 \end{bmatrix} \times \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\1 \\1 \end{bmatrix}$ and $N \cdot V = 1$.
For answer A., $$\widetilde N =  \begin{bmatrix} 
1 \\ -1 \\ 0 \end{bmatrix} \times \begin{bmatrix} 0 \\ -2 \\ 2 \end{bmatrix} = \begin{bmatrix} -2 \\-2 \\-2 \end{bmatrix} = (-2) N,$$ so condition 1 is satisfied. Moreover, $$N \cdot \widetilde V = \begin{bmatrix} 
1 \\ 1 \\ 1 \end{bmatrix} \cdot \begin{bmatrix} 
1 \\ 0 \\ 0 \end{bmatrix} = 0 = N \cdot V,$$ so condition 2 is satisfied. Therefore A. represents the same set as the solution set.
