The first equation in the Reduced Row Echelon Form tells you that we need $x_1+x_2=0$ and the second equation says $x_3=0$. So If we take $x_2=t$ for $t\in\mathbb{R}$ then we must have $x_1=-x_2=-t$ and $x_3=0$.
Thus we have a 1-dimensional solution space determined by the vector
$(x_1,x_2,x_3)=(-1,1,0)$.
This follows from the above discussion since taking $x_2=t$ corresponds to scalar multiplication by t on $(-1,1,0)$.
I became aware of this question by way of an answer on Meta and feel I must push back against the comment of DonAntonio and the answer of amWhy.
Uniqueness of the solution of the system
$$
\begin{aligned}
x_1&=3\\
x_2&=2\\
x_3&=3
\end{aligned}
$$
is obvious and needs no proof. What is there to prove? Is it conceivable that if you plug in numbers other than $3,$ $2,$ and $3$ for $x_1,$ $x_2,$ and $x_3$ you might obtain three true statements?
The determinant is a complicated object, and by bringing it in in this situation, you are making something simple appear much more difficult than it actually is.
Here's what I think you were probably getting at: Let's start with a simpler analogue. Is the solution of the equation $x=2$ unique? Of course is is: $2$ is the only solution. Now $x=2$ may be the end result of simplifying a more complicated equation, such as $13x=26.$ The latter is a special case of the general equation $ax=b.$ It is certainly the case that the latter has a unique solution if and only if $a\ne0.$ If $a=0,$ then there is no solution unless $b=0,$ in which case there are infinitely many solutions.
Likewise, the matrix equation $Ax=b,$ where $A$ is a square matrix and $x$ and $b$ are column vectors, has a unique solution if and only if $\det A\ne0.$ If $\det A=0,$ then it has either no solution or infinitely many solutions.
So it is helpful to introduce the determinant to make statements about the nature of the solution set of the general equation $Ax=b.$ But for concrete $A$ and $b,$ it is usually more efficient to row reduce the system than to compute $\det A.$ (More precisely, computing $\det A$ is best done by actually performing row reduction, but there is no need to mention determinants if you are row reducing to solve a concrete problem.) The end result of the row-reduction process will tell you whether there is a unique solution or not.
The only thing that might need proof is that the three row operations (swapping rows, multiplying a row by a non-zero number, adding a multiple of one row to another row) preserve the solution set. That is generally proved in a linear algebra course, and you can probably assume it from that point on. If not, let $S$ be a system and let $S'$ be the system that results from applying a row operation. You just need to prove that any solution to $S$ is a solution to $S',$ and that any solution to $S',$ is a solution to $S.$ This is straightforward, but it seems like overkill to do it in every row reduction problem you perform.
Best Answer
$$\lambda x_1+x_2+x_3=1$$ $$x_1+\lambda x_2+x_3=\lambda$$ $$x_1+x_2+\lambda x_3=\lambda^2$$ Adding all three equations together gives $$(\lambda+2)(x_1+x_2+x_3)=1+\lambda+\lambda^2$$ If $\lambda=-2$ we have $0=1+(-2)+(-2)^2=3$, which is absurd. Thus we can divide by $\lambda+2$:
$$x_1+x_2+x_3=\frac{1+\lambda+\lambda^2}{\lambda+2}$$ Subtracting this from all three original equations gives $$(\lambda-1)x_1=1-\frac{1+\lambda+\lambda^2}{\lambda+2}$$ $$(\lambda-1)x_2=\lambda-\frac{1+\lambda+\lambda^2}{\lambda+2}$$ $$(\lambda-1)x_3=\lambda^2-\frac{1+\lambda+\lambda^2}{\lambda+2}$$ If $\lambda=1$, all three of the original equations reduce to $x_1+x_2+x_3=1$, whose solution is simply $$x_1=p,x_2=q,x_3=1-p-q\qquad p,q\in\mathbb R\tag1$$ Otherwise, dividing by $\lambda-1$ gives the unique solution of $$x_1=\frac{1-\frac{1+\lambda+\lambda^2}{\lambda+2}}{\lambda-1}=\frac1{\lambda+2}-1\\ x_2=\frac{\lambda-\frac{1+\lambda+\lambda^2}{\lambda+2}}{\lambda-1}=\frac1{\lambda+2}\\ x_3=\frac{\lambda^2-\frac{1+\lambda+\lambda^2}{\lambda+2}}{\lambda-1}=\frac{(\lambda+1)^2}{\lambda+2}\tag2$$ In conclusion, the solutions of the linear system are