Hint: Any $2$ additional vectors will do, as long as the resulting $4$ vectors form a linearly independent set. Many choices! I would go for a couple of very simple vectors, check for linear independence. Or check that you can express the standard basis vectors as linear combinations of your $4$ vectors.
The null space of the matrix
$$\begin{bmatrix}
1 & 1 & -1 & -1 & 0 & 0 \\
1 & 1 & 0 & 0 & -1 & -1 \\
0 & 0 & 1 & 1 & -1 & -1 \\
\end{bmatrix}$$
is, by definition, the set of vectors $\mathbf{x}=(x_1,x_2,x_3,x_4,x_5,x_6)$ such that
\begin{align*}
x_1 + x_2 - x_3 - x_4 &= 0 \\
x_1 + x_2 - x_5 - x_6 &= 0 \\
x_3 + x_4 - x_5 - x_6 &= 0
\end{align*}
So, this is the vector space $V$ in question. The matrix has rank $2$, so the dimension of $V$ is $4$ by the Rank Nullity Theorem.
Thus, any $4$ linearly independent vectors in $V$ form a basis. A natural try is $$\{( 1, 1, 1, 1, 1, 1 ),(0, 0, 0, 0, 1, -1),( 0, 0, 1, -1, 0, 0 ),(1, -1, 0, 0, 0, 0)\}.$$ To prove that it is indeed a basis, we check that the matrix $$\begin{bmatrix}
1 & 1 & 1 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 1 & -1 \\
0 & 0 & 1 & -1 & 0 & 0 \\
1 & -1 & 0 & 0 & 0 & 0 \\
\end{bmatrix}$$ has rank $4$ (e.g. by performing Gaussian elimination).
The column space of the transpose of the above matrix, i.e.
$$\begin{bmatrix}
1 & 0 & 0 & 1 \\
1 & 0 & 0 & -1 \\
1 & 0 & 1 & 0 \\
1 & 0 & -1 & 0 \\
1 & 1 & 0 & 0 \\
1 & -1 & 0 & 0 \\
\end{bmatrix},$$
will be $V$ (or, at least, will be isomorphic to $V$: all the vectors will be transposed); see Wikipedia.
Best Answer
The sum of the two equations gives
$$x_3=4x_2-3x_1$$
the second one yields to
$$x_4=x_1-x_2+2x_3$$ $$=x_1-x_2+2(4x_2-3x_1)$$ $$=-5x_1+7x_2$$
thus
$$(x_1,x_2,x_3,x_4)\in V\iff$$ $$(x_1,x_2,x_3,x_4)=$$ $$(x_1,x_2,-3x_1+4x_2,-5x_1+7x_2)=$$ $$x_1(1,0,-3,-5)+x_2(0,1,4,7)=$$ $$x_1\vec{u}+x_2\vec{v}$$
$\vec{u} $ and $ \vec{v} $ are clearly independent, so $ (\vec{u},\vec{v}) $ is a basis of $ V$ and dim$(V)=2$.