Expressing a projective conic in standard form I was given the task to find a projective transformation $\phi: \mathbb{P}_\mathbb{C} \rightarrow \mathbb{P}_\mathbb{C}$ that transforms the following conic:
$$2\,{x}^{2}+2\,xy-3\,xz+4\,yz+{z}^{2}=0$$
into the form:
$$X^2 + Y^2 + Z^2 = 0$$
Then my attempt to do so was to first find the matrix of the symmetric bilinear form which is given by:
$$ \left[ \begin {array}{ccc} 2&1&-3/2\\ 1&0&2
\\ -3/2&2&1\end {array} \right] 
$$
Then I wanted to orthogonally diagonalise this matrix to obtain an expression that doesn't have any cross terms in it. However the eigenvectors of this matrix are not "nice" (I've checked with Maple and they involve a lot of square roots).
In the solution given to me, the matrix that expresses this conic in standard form was given by the following matrix:
$$  \left[ \begin {array}{ccc} 1/2\,\sqrt {2}&1/10\,\sqrt {10}&-1/3\,
\sqrt {3}\\ 0&1/10\,\sqrt {10}&-i/6\sqrt {3}
\\ 0&1/5\,\sqrt {10}&i/3\sqrt {3}\end {array}
 \right] 
$$
I can see that this matrix is orthogonal, but how does it simplify the expression of the conic above? And how do I find it from scratch? And how do this in general?

I found some notes online, the task is to begin with symmetric $H$ and solve $P^T HP = D$ diagonal, which can be done over the reals. Then forcing the identity over the complexes requires multiplying by an additional diagonal complex matrix on both sides.
Algorithm discussed at reference for linear algebra books that teach reverse Hermite method for symmetric matrices
$$ P^T H P = D  $$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
 -  \frac{ 1 }{ 2 }  & 1 & 0 \\ 
 - 2 &  \frac{ 11 }{ 2 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
4 & 2 &  - 3 \\ 
2 & 0 & 4 \\ 
 - 3 & 4 & 2 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 &  -  \frac{ 1 }{ 2 }  &  - 2 \\ 
0 & 1 &  \frac{ 11 }{ 2 }  \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
4 & 0 & 0 \\ 
0 &  - 1 & 0 \\ 
0 & 0 & 30 \\ 
\end{array}
\right) 
$$
$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
 \frac{ 1 }{ 2 }  & 1 & 0 \\ 
 -  \frac{ 3 }{ 4 }  &  -  \frac{ 11 }{ 2 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
4 & 0 & 0 \\ 
0 &  - 1 & 0 \\ 
0 & 0 & 30 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 &  \frac{ 1 }{ 2 }  &  -  \frac{ 3 }{ 4 }  \\ 
0 & 1 &  -  \frac{ 11 }{ 2 }  \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
4 & 2 &  - 3 \\ 
2 & 0 & 4 \\ 
 - 3 & 4 & 2 \\ 
\end{array}
\right) 
$$
Alright, given this $P$ and the resulting $D,$  we multiply on the right and left by
$$
P_1 = \left( 
\begin{array}{rrr} 
\frac{1}{2} & 0 &  0 \\ 
0 & i & 0 \\ 
 0 & 0 & \frac{1}{\sqrt {30}} \\ 
\end{array}
\right) 
$$
to reach $I.$ I began with double your symmetric matrix, so you may need to multiply throughout by $\sqrt 2$ to take your actual matrix to $I.$ Note how completely the answer changes based on choices made; mine looks nothing like your given answer.
