Here, subspace $X \subseteq \mathbb{R}^{3}$ is defined as :-

$$X = \left \{x \in \mathbb{R}^3 \mid x_1 + x_2 + x_3 = 0, \text{ where } x^T = \left[x_1,x_2,x_3\right]^T\right \}$$.

which means sum of all the components of the vectors are zero which are in subspace $X$.

We can write $ x_1 + x_2 + x_3 = 0$ as

$\begin{bmatrix} 1 &1 &1 \end{bmatrix}$$\begin{bmatrix} x_{1}\\x_2 \\ x_3 \end{bmatrix} = 0$

Now, assuming vector $x$ is in the nullspace of matrix $[1,1,1]$ and we have to find the basis for it.

Since, all the components of vector $x$ are zero. So, we can write $x_3 = -x_2 - x_1$

So,

$\begin{bmatrix} x_{1}\\x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} x_{1}\\x_2 \\ -x_2-x_1 \end{bmatrix} =x_1\begin{bmatrix} 1\\0 \\ -1 \end{bmatrix} + x_2\begin{bmatrix} 0\\1 \\-1 \end{bmatrix}$ where $x_1,x_2 \in \mathbb{R}$

So, vectors $\left \{ \begin{bmatrix} 1\\0 \\-1 \end{bmatrix}, \begin{bmatrix} 0\\1 \\-1 \end{bmatrix} \right \}$ should be in the subspace $X.$

Now, we have to check whether these vectors are linearly independent or not.

So,

$k_1\begin{bmatrix} 1\\0 \\-1 \end{bmatrix} + k_2\begin{bmatrix} 0\\1 \\-1 \end{bmatrix} = \begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

$\Rightarrow$ $k_1 = 0$ and $k_2 = 0$ which implies both vectors $\left \{ \begin{bmatrix} 1\\0 \\-1 \end{bmatrix}, \begin{bmatrix} 0\\1 \\-1 \end{bmatrix} \right \}$ are linearly independent.

Since, $x \in \mathbb{R}^{3}$ and it is in nullspace which implies no. of columns($n$) = $3$ (it can be seen from the matrix $[1,1,1])$.

Now, say, matrix $[1,1,1]$ is $A$. So, $rank(A)= r = 1$.

So, $dimension(N(A)) = n-r = 3-1 =2 $. It means basis for nullspace should have $2$ vectors. Basis contains linearly independent vectors which must span the whole vector space.

**Hence, we can say that here basis has $2$ linearly independent vectors which are $\left \{ \begin{bmatrix} 1\\0 \\-1 \end{bmatrix}, \begin{bmatrix} 0\\1 \\-1 \end{bmatrix} \right \}$**

**So, Basis for $X$ is $\left \{ \begin{bmatrix} 1\\0 \\-1 \end{bmatrix}, \begin{bmatrix} 0\\1 \\-1 \end{bmatrix} \right \}.$**

**Hence, Answer is $(A)$**