Gram-Schmidt orthornormalization , $\dim(S)$. By means of Gram-Schmidt orthonormalization, find an orthonormal basis in 
$$S=\{v=\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{pmatrix}: x_{1}-x_{2}+2x_{3}-3x_{4}=0\}$$
subspace of $\left(\mathbb{R}^{4},\langle\,,\,\rangle\right)$.
What I do: 
$x_{2}=\alpha, x_{3}=\beta, x_{4}=\gamma$ so $x_{1}=\alpha-2\beta+3\gamma$ and :
$$\left(\alpha-2\beta+3\gamma,\alpha,\beta,\gamma\right)=\left(\alpha, \alpha,0,0\right)+\left(-2\beta,0, \beta,0\right)+\left(3\gamma,0,0,\gamma\right).$$
So a base is : $\{(1,1,0,0),(-2,0,1,0),(3,0,0,1)\}$ and I have to apply Gram-Schmidt for this base? 
Another question, what is the $\dim(S)$? $3$ or $4$ ? 
because I can find canonical base in $\mathbb{R}^{4}$ to write $\left(\alpha-2\beta+3\gamma,\alpha,\beta,\gamma\right)$.
Thanks :) 
 A: The question was basically answered in comments, so I am posting a CW-answer, so that it is not left unanswered.
Since you exhibited a basis for $S$ which has 3 elements, you get that $\operatorname{dim}(S)=3$. Using Gram-Schmidt process  you will obtain from this basis an orthonormal basis of the same subspace.
If you have any doubts whether your three vectors $(1,1,0,0)$, $(-2,0,1,0)$ and $(3,0,0,1)$ are linearly independent, you just need to have a look at the last three coordinates. If $c_1(1,1,0,0)+c_2(-2,0,1,0)+c_3(3,0,0,1)=(\dots,c_1,c_2,c_3)=(0,0,0,0)$, then obviously $c_1=c_2=c_3=0$. 
This is similar to the argument which is used to show that non-zero rows of matrix in the reduced row echelon form are linearly independent. For a slightly more general claim, see this question.
A: So the orthonormal basis vectors of the nullspace given in this post are as follows (using the Gram-Schmidt process):
\begin{eqnarray}
q_1^T &=& (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \, 0, \, 0)\, , \\
q_2^T &=& (\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} , 0)\, , \\
q_3^T &=& (\frac{1}{\sqrt{10}}, \frac{-1}{\sqrt{10}}, \frac{2}{\sqrt{10}} , \frac{2}{\sqrt{10}})\, .
\end{eqnarray}
