Your vectors are neither orthogonal ($x_2 \cdot x_3 = -1.5)$, nor normal.
The note to your question points to the choice, given a normal vector $n$, of choosing either $n$ or $-n$ as normal.
How to orthonormalize:
We could start with $x_1$ and keep it as first vector of the new base, $x_1' = x_1$.
We want to end up with normal vectors as well, so it is better to do the normalization now.
The second vector is the second given vector without the parts in direction of the first new base vector:
\begin{align}
x_1''
&= \frac{x_1'}{\sqrt{x_1' \cdot x_1'}} = \frac{x_1}{\sqrt{x_1 \cdot x_1}} \\
x_2' &= x_2 - (x_1'' \cdot x_2) x_1'' \\
x_2'' &= \frac{x_2'}{\sqrt{x_2' \cdot x_2'}} \\
x_3' &= x_3 - (x_1'' \cdot x_3) x_1''- (x_2'' \cdot x_3) x_2'' \\
x_3'' &= \frac{x_3'}{\sqrt{x_3' \cdot x_3'}} \\
\end{align}
Thus here
\begin{align}
x_1'' &= (0, 1/\sqrt{2}, 1/\sqrt{2}) \\
x_2' &= (3, -3/2, 3/2) \\
x_2'' &= \sqrt{\frac{2}{27}}(3,-3/2, 3/2) \\
x_3' &= (2/3, 2/3, -2/3) \\
x_3'' &= \frac{\sqrt{3}}{2} (2/3, 2/3,-2/3) = (1/\sqrt{3},1/\sqrt{3},-1/\sqrt{3})
\end{align}
For the third vector we removed the parts in direction of the new first and second vector.
Only $x_3''$ has negative third component, so we choose
$$
x_3''' = - x_3'' = (-1/\sqrt{3},-1/\sqrt{3},/\sqrt{3})
$$
You can start the Gram-Schmidt process with an arbitrary basis whose first element is $w_1$. It's easy to see that such a thing exists. You can construct it as follows. At each step you'll have a set of linearly independent vectors. As long as there are fewer of them than the dimension of your space, take any vector that is not in the linear span of your set, and insert it into the set.
Best Answer
Please, be more specific when posting the question. What exactly are your matrices $Q$, $R$ and $A$ that you talk about?
One way of checking would be to write down your basis as columns of a matrix, let's call it $M$. Then you simply have to verify that $M^{\mathrm{T}}M=\mathrm{id}$ holds. This would mean that the columns form an ONB.