Extract a maximal independent subsystem and complete this to a basis in $\mathbb{R}^3$ 
I have $4$ vectors $v_1=(0,-1,2),$ $v_2= (1,0,1),$ $v_3= (-3,-2,1),$ $v_4= (1,-2,5)$ which form a dependent system and $S=\text{span}\{v_1,v_2,v_3,v_4\}.$ Find a maximal independent subsystem of $S$ and complete this subsystem to a basis in $\mathbb{R}^3$.

My idea is to find a maximal subsystem is to find $3$ vectors and to prove that the determinant isn't $0.$ But can $4$ vectors form a basis in $\mathbb{R}^3?$
 A: Note that $(1,0,1) = (1,-2,5) - 2\cdot(0,-1,2)$ so $v_2 = v_4-2\cdot v_1.$
Therefore $$\langle v_1,v_2,v_3,v_4 \rangle = \langle v_1,v_3,v_4 \rangle.$$
But $(0,-1,2) = \frac{1}{8}((-3,-2,1) + 3\cdot(1,-2,5)),$ so $v_1 = \frac{1}{8}(v_3 + 3\cdot v_4).$
Hence
$$\langle v_1,v_3,v_4 \rangle = \langle v_3,v_4 \rangle.$$
Since $v_3 \neq \lambda \cdot v_4, \forall \lambda \in \mathbb{R},$ we conclude that $A = \{v_3,v_4\}$ is linearly independent.
To form a basis of $\mathbb{R}^{3},$ first note the $\dim{\mathbb{R}^3} = 3$ but $\dim{A} = 2.$
So $A$ itself doesn’t form a basis of $\mathbb{R}^3.$ But (by Steinitz Theorem) we can form a basis of $\mathbb{R}^3$ that contains $A$ by adding one more vector.
But this extra vector has to be chosen carefully, note that this one, when reunited with $A$ must “preserve” the linearly independence of $A.$
We can pick this vector, for example, from the standard basis of $\mathbb{R}^3,$ which is
$$((1,0,0),(0,1,0),(0,0,1)).$$
Take, for example, $(1,0,0).$ Then
$$((-3,-2,1),(1,-2,5),(1,0,0))$$
form a basis of $\mathbb{R}^3.$
