The most generic answer: any time that we can reduce a problem over an incredibly general object (say, a matrix) to a problem in which we have more information at our fingertips (say, the same problem but over matrices that are in JCF), we make life easier - both in terms of proving theory and in terms of practical computations.
To be more specific to the situation at hand: the Jordan canonical form is sort of the next-best-thing to diagonalization. If the matrix is diagonalizable, then its JCF is diagonal; if it isn't, then what you get is at least block diagonal, and the blocks come in a predictable form.
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 have
\begin{align} Q(x_1,x_2,x_3)&=x^2_1+x^2_2+2x_1x_2-2x_1x_3-6x_2x_3 \\&=\color{red}{x_1^2+2x_1(x_2-x_3)+(x_2-x_3)^2}\color{blue}{-(x_2-x_3)^2+x_2^2-6x_2x_3} \\&=\color{red}{(x_1+x_2-x_3)^2}\color{blue}{-\left(x_3^2+2x_3\cdot2x_2+4x_2^2-4x_2^2\right)} \\&=(x_1+x_2-x_3)^2-(x_3+2x_2)^2+4x_2^2 \end{align}
So $Q(x_1,x_2,x_3)$ is transformed to some
$$Q(y_1,y_2,y_3)=y_1^2-y_2^2+y_3^2\,,$$
where
$$\begin{pmatrix}y_1\\y_2 \\y_3 \\\end{pmatrix}=\begin{pmatrix}1&1&-1\\0&2&1 \\0&2&0 \\\end{pmatrix}\begin{pmatrix}x_1\\x_2 \\x_3 \\\end{pmatrix}$$
is the nonsingular transformation.