Original matrix A is:
$$\begin{pmatrix} 1 & 2 & 7& 1\\ 1 & 1 & 2& 0\\ 2&0&-6&-2 & \\ 1&1&2&0 \end{pmatrix}$$
$w=(A*\begin{pmatrix} x1\\ x2\\ x3\\ x4 \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 0\\ 0 \end{pmatrix})$ Find base set of vector for W.
Using Gaus method I get:
$$\begin{pmatrix} 1 & 2 & 7&1\\ 0&-1&-5&-1 \\ 0 & 0 & 0&0\\ 0&0&0&0 \end{pmatrix}$$ which has 2 leading elements. So the column of each elemnt suppose to be part of base, these are: $(1,1,2,1)$ , $(2,1,0,1)$ but seemingly it's wrong because the answer is written like this: ${(3,-5,1,0),(1,-1,0,1)}$
where is my mistake?
Take your reduced system
$$\begin{pmatrix}1 &2 &7& 1\\0 &-1 &-5 &-1\\ 0 &0 &0 &0\\ 0 &0 &0 &0\end{pmatrix}\begin{pmatrix}x\\y\\z\\w\end{pmatrix}=\begin{pmatrix}x+2y+7z+w\\-y-5z-w\\0\\0\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}$$
You now just have to solve the equation for $v=(x,y,z,w)^\top$. Working with the method of back-substitution, we first assign all the non-pivots(the variables not leading any row) a real constant, as they may vary in their value arbitrarily and we then can determine the pivot variables in dependence on them.
Thus, let $w=c,z=c'$(for constants $c,c'$). Then, by the second row of the vector equation, we may determine the value of $y$ in terms of the constants $c,c'$(values of $z,w$):
$$-y-5z-w=0\Leftrightarrow-y-5c'-c=0\Leftrightarrow y=-5c'-c$$
Continuing back-substitution, we now derive from the first row the value of $x$ in dependence of $c,c'$:
$$x+2y+7z+w=0\Leftrightarrow x=-2y-7z-w\Leftrightarrow x=10c'+2c-7c'-c\Leftrightarrow x=3c'+c$$
Thus, your solutions are any vectors of the following form
$$(3c'+c,-5c'-c,c',c)=c'(3,-5,1,0)+c(1,-1,0,1)$$
for arbitrary $c,c'\in\mathbb{R}$.
The solution set of $Av=\mathbf0$, i.e. the set $S(A,\mathbf0)=\{v\in\mathbb{R}^4\mid Av=\mathbf0\}$, for which you want to determine a basis, is commonly called the kernel of the matrix $A$, written $\mathrm{ker}(A)$.
By the above, we've found that any solution to $Av=\mathbf0$, i.e. any $v\in\mathrm{ker}(A)$, can be written as
$$v=c'(3,-5,1,0)+c(1,-1,0,1)$$
Thus, in other terminology, we say that any $v\in\mathrm{ker}(A)$ is a linear combination of $(3,-5,1,0)$ and $(1,-1,0,1)$. The set of all linear combination for a set of vectors $S$ is called the span(or linear hull) of the set $S$ and I denote this with $\mathrm{span}(S)$.
We may thus rephrase our result about the solutions of $\mathrm{ker}(A)$ as such, that
$$\mathrm{ker}(A)=\mathrm{span}(\{(3,-5,1,0),(1,-1,0,1)\})$$
A basis, for any space $V$, is defined as linearly independent set of vectors spanning $V$, i.e. a linearly independent set $B$, s.t. $\mathrm{span}(B)=V$.
Now, as $(3,-5,1,0),(1,-1,0,1)$ are linearly independent(check this yourself), it follows that they also constitute a basis for this set(space).
Maybe, on a last note, try to generalize the process I've just outline. Toying with these notions and especially with the connections between subspaces, linear systems of equation and matrices in combination is a great exercise. E.g. show that for any $n\times n$ real matrix $A$, $\mathrm{ker}(A)$ always forms a subspace. Is this true for any solution set $S(A,b)$ for a vector equation $Av=b$?