Creating an orthonormal basis 
Let there be $W=Span\{(1,0,1),(1,2,3)\}$ sub-space of $V^3$ find an orthonormal basis for $V^3$ 

So at first I have found 2 orthonormal vectors $e_1=\frac{1}{\sqrt{2}}(1,0,-1)$ and $e_2=\frac{1}{\sqrt{3}}(1,1,1)$, which vector can I take to create the third orthonormal vector? 
 A: You can take ${1\over\sqrt{6}}(1,-2,1)$.
Edited:
I want to find a vector orthogonal to $(1,0,-1)$ and also to $(1,1,1)$. I know that if $v$ is any vector orthogonal to these two, then the three of them together form an orthogonal basis to $\mathbb{R}^3$.
So assume $(x,y,z)\in\mathbb{R}^3$ is orthogonal to $(1,0,-1)$. Then :
$$x\cdot 1+y\cdot 0 + y\cdot (-1) = 0$$
therefore, if $(x,y,z)$ is orthogonal to $(1,0,-1)$, we must have 
$$x = z$$
Now, among all vectors $(x,y,z)$ that are orthogonal to $(1,0,-1)$ I want to find one orthogonal to $(1,1,1)$ too. Now if $(x,y,z)$ is orthogonal to $(1,1,1)$ then
$$x\cdot 1 + y\cdot 1 + z\cdot 1 = 0$$
so 
$$x + y + z = 0$$
and I already know that $x=z$ so
$$x + y + x = 0$$
so
$$2x = -y$$
So far I have shown that if $(x,y,z)$ is orthogonal to both the vectors $(1,0,-1)$ and $(1,1,1)$ then necessarily $x=z$ and $2x=-y$. This means that once we know $x$, we also know $y$ and $z$, so the vector has the form $(x,-2x,x)$. Conversely, every vector $(x,y,z)$ of the form $(x,-2x,x)$ is orthogonal to both the vectors $(1,0,-1)$ and $(1,1,1)$, as can be verified by calculating the scalar product of $(x,-2x,x)$ with the vectors $(1,0,-1)$ and $(1,1,1)$. For example, 
$$(x,-2x,x)\cdot (1,1,1) = x-2x+x = 0$$
I leave the verification that $(x,-2x,x)$ is orthogonal to $(1,0,-1)$ to the readers.
So now I know that every choice of $x$ gives me a vector $(x,-2x,x)$ orthogonal to both the vectors above. I therefore choose $x=1$ and obtain the vector $(1,-2,1)$. I now compute it's norm and find that it is $\sqrt{6}$, so the normalized vector is ${1\over\sqrt{6}}(1,-2,1)$.
No need for Gram Schmidt here.
A: Compute the cross product
$e_1 \times e_2 = \dfrac{1}{\sqrt 6} (1,0,-1) \times (1,1,1) = 
\dfrac{1}{\sqrt 6}(1, -2, 1)$
