Basis for a eigenspace (multiple choice problem) The following (multiple choice) problem is from a test review.

For the given matrix $A$, find a basis for the corresponding eigenspace for the given eigenvalue.
$$A = \begin{bmatrix}1 & 6 & 6 \\ 6 & 1 & -6 \\ -6 & 6 & 13\end{bmatrix};\quad \lambda = 7.$$

The four given options are

A) $\left\{ \begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ -1 \end{bmatrix} \right\}$
B) $\left\{ \begin{bmatrix}0 \\ 1 \\ -1 \end{bmatrix} \right\}$
C) $\left\{ \begin{bmatrix}1 \\ 0 \\ -1 \end{bmatrix} \right\}$
D) $\left\{ \begin{bmatrix}1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix} \right\}$

The correct answer is A.

My answer is not among the answer choices.  What am I doing wrong?
First, reduce $(A-\lambda I)$:
$$\begin{align}
A - \lambda I &=
\begin{bmatrix}1 & 6 & 6 \\ 6 & 1 & -6 \\ -6 & 6 & 13\end{bmatrix} - \begin{bmatrix}7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7\end{bmatrix} \\
&= \begin{bmatrix}-6 & 6 & 6 \\ 6 & -6 & -6 \\ -6 & 6 & 6\end{bmatrix} \\
&\sim \begin{bmatrix}-1 & 1 & 1 \\ 1 & -1 & -1 \\ -1 & 1 & 1\end{bmatrix} \\
&\sim \begin{bmatrix}1 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}
\end{align}$$
Parametric form solving for $(A-\lambda I) = \vec 0$:
$$\begin{align}
x_1 &= x_2 + x_3 \\
x_2 &= x_2 \\
x_3 &= x_3
\end{align}$$
has solutions $x_2(1,1,0) + x_3(1,0,1)$ for all $x_2 , x_3$.
So a basis for the null-space of $(A-\lambda I)$ consists of the vectors $\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$ and $\begin{bmatrix}1 \\0 \\ 1\end{bmatrix}$.
So, am I wrong or is the solution wrong?
 A: As @GTonyJacobs points out in the comments, you are correct but the key is also correct. Vector spaces can have many different bases!
Note that you have proven that $A-7I$ has rank one so the rank-nullity theorem implies that the nullspace of $A-7I$ has dimension two. This allows us to immediately eliminate choices (B) and (C). 
Now, note that
\begin{align*}
A\begin{bmatrix}1\\0\\1\end{bmatrix}
&= \begin{bmatrix}7\\0\\7\end{bmatrix}
= 7\begin{bmatrix}1\\0\\1\end{bmatrix}
&
A\begin{bmatrix}0\\1\\-1\end{bmatrix}
&=\begin{bmatrix}0\\7\\-7\end{bmatrix}
=7\begin{bmatrix}0\\1\\-1\end{bmatrix}
\end{align*}
This proves that choice (A) is correct. 
Also, note that choice (D) is incorrect since
$$
A\begin{bmatrix}1\\0\\-1\end{bmatrix}
=\begin{bmatrix}-5\\12\\-19\end{bmatrix}
\neq7\begin{bmatrix}1\\0\\-1\end{bmatrix}
$$
A: As @GTonyJacobs correctly said in a comment, your answer is correct, but there are infinitely many answers to the question and the one you found is not the one given as one of the multiple choice answers.
You asked @GTonyJacobs in a comment of your own what to do in such a case. The first thing is for your to recognize if a question has multiple answers. Any question asking for a basis of a non-trivial vector space over the rationals or reals will have infinitely many answers. If your given basis is not one of the choices given, you then need to see which of the given answers spans the same space as your basis.
In this particular case, this is easily done. One of your vectors, $\begin{bmatrix}1\\0\\1\end{bmatrix}$, is one of the vectors in choice (A). If you add the two vectors in choice (A) you get your other basis vector, $\begin{bmatrix}1\\1\\0\end{bmatrix}$. Adding two vectors is an elementary operation, so the two bases cover the same subspace.
In a more difficult case, write each basis as a matrix of row vectors then calculate the reduced row echelon form. The two subspaces covered by the bases are equal if and only if the two reduced row echelon forms are equal. In this particular case, you want to reduce the two matrices
$$\begin{bmatrix}1&0&1\\0&1&-1\end{bmatrix},\quad \begin{bmatrix}1&1&0\\1&0&1\end{bmatrix}$$
(Choice (A) is first, yours is second.) These both reduce to
$$\begin{bmatrix}1&0&1\\0&1&-1\end{bmatrix}$$
which is choice (A), of course, though that is not relevant. This shows that your basis and choice (A) cover the same subspace, so you choose (A).
By the way, choices (B) and (C) can be rejected immediately, since they have the wrong number of dimensions. You could find the reduced row echelon form of choice (D) and see that it differs from yours.
