Your reduced row echelon form is $\left(\begin{array}{ccc} 1 & 1 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right).$ You have one leading variable (the first term), and two free variables. The system is equivalent to the single equation $x+y+z=0.$ That means that you have two degrees of freedom, $y$ and $z$. You can express $x$ in terms of $y$ and $z$ and $x=-y-z$. So the solutions are given by: $\begin{align*} x &= -s - t\\ y & = s\\ z &= t \end{align*}\qquad\qquad s,t\in\mathbb{R}.$ You get a basis for the space of solutions by taking the parameters (in this case, $s$ and $t$), and putting one of them equal to $1$ and the rest to $0$, one at a time. Setting $s=1$ and $t=0$, we get $x=-1$, $y=1$, $z=0$, leading to the vector $(-1,1,0)$; setting $s=0$ and $t=1$ we get $x=-1$, $y=0$, $z=1$, leading to the vector $(-1,0,1)$.
Note: What you claim is "the solution" is just one possible basis for the eigenspace. The eigenspace is equivalent to the nullspace of $A-\lambda I$; which is equivalent to the nullspace of the reduced row echelon form of $A-\lambda I$. So this question really comes down to:
Do you know how to find a basis for the nullspace of a matrix $A$?
Equivalently:
Do you know how to express the set of solutions to a homogeneous system of equations in vector form?
Above, we end up with the single-equation "system" given by $x+y+z=0.$ Do you know how to write down/describe the set of all solutions to this equation in terms of vectors?