Method 2 is just an algorithm to find a diagonal matrix which is congruent to the first, so maybe you can call it matrix congruence method.
Another method along the lines of matrices is to do orthogonal diagonalization, but method 2 would be faster...
You can do all the row operations first: but be careful then you have $P^TA$, so the safest then is to take the resulting matrix on the right, in your augmented matrix (which is $P^T$), transpose it and multiply on the right which gives you $P^TAP$ - I don't think you will gain too much when you do this - it is much simpler to just alternate row and column operations.
Completing squares I get, with $\;x,y,z\;$ instead of $\;x_i\;$ :
$$3x^2-2xy+2y^2-2yz+3z^2=3\left(x-\frac13y\right)^2+\frac43y^2+3\left(z-\frac13y\right)^2\implies$$
$$\implies \begin{cases}I\;\;\;\;\,a=x-\frac13y\\II\;\;\;b=y\\III\;c=z-\frac13y\end{cases}\;\;\;
\implies\;\;\begin{cases}x=a+\frac13b\\y=b\\z=c+\frac13b\end{cases}$$
So the base part is $\;\left(1,\,\frac13,\,0\right)\;,\;(0,1,0)\;,\;\left(0,\,\frac13,\,1\right)\;$ , and the quadratic can be expressed as
$$\color{red}{q(a,b,c)=3a^2+\frac43b^2+3c^2}$$
Observe that the eigenvalues of $\;A\;$ are
$$\det(xI-A)=\begin{vmatrix}x-3&1&0\\1&x-2&1\\0&1&x-3\end{vmatrix}=$$
$$=(x-2)(x-3)^2-2(x-3)=(x-3)(x-4)(x-1)$$
with eigenvectors:
$$\begin{align*}
&\bullet\;\lambda=1:\;\;\begin{cases}-2x+y\;\;\;\;\;\;\;=0\\\;\;\;\;\;\;\;\;\;\;y-2z=0\end{cases}\implies2x=2z=y\implies&\begin{pmatrix}1\\2\\1\end{pmatrix}\\{}
&\bullet\;\lambda=3:\;\;\begin{cases}\;\;\;\;\;\;y\;\;\;\,\;\;=0\\x+y+z=0\end{cases}\implies x=-z\,,\,y=0\implies&\begin{pmatrix}1\\0\\\!-1\end{pmatrix}\\{}
&\bullet\;\lambda=4:\;\;\begin{cases}x+y\;\;\;\;\;\;=0\\\;\;\;\;\;\,y+z=0\end{cases}\implies x= z=-y\implies&\begin{pmatrix}\;1\\\!-1\\\;1\end{pmatrix}\end{align*}$$
Now orthonormalize the above three vectors, form the matrix $\;P\;$ whose columns are the new vectors and check that
$$P^{-1}AP=\begin{pmatrix}1&0&0\\0&3&0\\0&0&4\end{pmatrix}$$
Best Answer
You're quite right that there cannot be a cubic term in a quadratic form -- for example, as written $Q(0,0,2)\ne 4Q(0,0,1)$, contradicting the definition.
It looks to me like the apparently cubic term is just a typo in the problem. It should be $x_3^2$, not $x_3^3$.
If you are to hand in written homework, I would simply start it with: