Determine a Basis of the vector subspace of $\mathbb{R^4}$ spanned by given vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ I would like your feedback for the below problem

Given are the vectors
$$\vec{a} = (-2, 3, 1, -1)^T$$
$$\vec{b} = (0, 1, -1, 0)^T$$
$$\vec{c} = (6, 2, 0, 3)^T$$
$$\vec{d} = (6, 0, 2, 3)^T$$
Determine a Basis of the vector subspace of $\mathbb{R^4}$, which is spanned by those vectors.

To find a Basis, as far as I know, we just need normalized linearly independent vectors. So first, we check if they are linearly independent using Gaussian elimination. 
 
Setting the vector combination equal to the vector $(0, 0, 0, 0)^T$ and using for example https://matrixcalc.org/en/ ,
( https://matrixcalc.org/en/slu.html#solve-using-Gaussian-elimination(%7B%7B-2,0,6,6,0%7D,%7B3,1,2,0,0%7D,%7B1,-1,0,2,0%7D,%7B-1,0,3,3,0%7D%7D) )
it turns out that the last row can be zeroed out, so the coefficient of vector $\vec{d}$ is a free variable. Solving further we see that the coefficients of $\vec{b}$ and $\vec{c}$ can be expressed with the coefficent of $\vec{d}$, in our case $a = 0, b = 2d, c = -d, d = d$
So a basis of a vector subspace of $\mathbb{R}^4$ could be vectors $\vec{a}, \vec{b}, \vec{c}$ but not vector $\vec{d}$ because he is linearly dependent.
We still need to normalize the vectors. We can just divide each vector by the square root of the sum of each component squared. So
$$\vec{a}_{basis} = \frac{\vec{a}}{\sqrt{15}}$$
$$\vec{b}_{basis} = \frac{\vec{b}}{\sqrt{2}}$$
$$\vec{c}_{basis} = \frac{\vec{c}}{\sqrt{49}} = \frac{\vec{c}}{7}$$
So those vectors form a basis of the vector subspace of $\mathbb{R}^4$

Is this correct ?
Thanks for your support !
 A: You don't need to normalize anything. You simply reduce it to a Row Reduced Echelon form(which is very similar to the process of Gaussian Elimination) to get a matrix whose non-zero rows are linearly independent and they will give you your basis.
$$\begin{bmatrix} -2&3&1&-1\\ 0&1&-1&0 \\ 6&2&0&3 \\ 6&0&2&3 \end{bmatrix}$$
Apply $R_{3}=R_{3}+3R_{1} ,\, R_{4}=R_{4}+3R_{1}$ to get
$$\begin{bmatrix} -2&3&1&-1\\ 0&1&-1&0 \\ 0&11&3&0 \\ 0&9&5&0 \end{bmatrix}$$
Apply $R_{1}=R_{1}-3R_{2},\,R_{3}=R_{3}-11R_{2}\, , R_{4}= R_{4}-9R_{2}$ to get
$$\begin{bmatrix} -2&0&4&-1\\ 0&1&-1&0 \\ 0&0&14&0 \\ 0&0&14&0 \end{bmatrix}$$
Apply $R_{3}=\frac{R_{3}}{14}\,, R_{4}=\frac{R_{4}}{14}$ to get :-
$$\begin{bmatrix} -2&0&4&-1\\ 0&1&-1&0 \\ 0&0&1&0 \\ 0&0&1&0 \end{bmatrix}$$
Apply $R_{1}=R_{1}-4R_{3} \, , R_{2}=R_{2}+R_{3} \, , R_{4}=R_{4}-R_{3}$ to get
$$\begin{bmatrix} -2&0&0&-1\\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&0 \end{bmatrix}$$
This gives you that $\{(-2,0,0,-1)^{T},(0,1,0,0)^{T},(0,0,1,0)^{T}\}$ is a basis for the row space(which is equal to the subspace spanned by the vectors which we took as row vectors).
Which you can also write as $\{(2,0,0,1),(0,1,0,0),(0,0,1,0)\}$ if you don't want negative sign.
