4x^2-12xy+9y^2+z^2+4xz-6yz+4x-6y+2z-5=0 canonical form

Given equation of the surface of 2-order:
$$4 x^{2} - 12 x y + 4 x z + 4 x + 9 y^{2} - 6 y z - 6 y + z^{2} + 2 z - 5 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 4$$
$$a_{12} = -6$$
$$a_{13} = 2$$
$$a_{14} = 2$$
$$a_{22} = 9$$
$$a_{23} = -3$$
$$a_{24} = -3$$
$$a_{33} = 1$$
$$a_{34} = 1$$
$$a_{44} = -5$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$

     |a11  a12|   |a22  a23|   |a11  a13|
I2 = |        | + |        | + |        |
     |a12  a22|   |a23  a33|   |a13  a33|

$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$

     |a11  a14|   |a22  a24|   |a33  a34|
K2 = |        | + |        | + |        |
     |a14  a44|   |a24  a44|   |a34  a44|

     |a11  a12  a14|   |a22  a23  a24|   |a11  a13  a14|
     |             |   |             |   |             |
K3 = |a12  a22  a24| + |a23  a33  a34| + |a13  a33  a34|
     |             |   |             |   |             |
     |a14  a24  a44|   |a24  a34  a44|   |a14  a34  a44|

substitute coefficients
$$I_{1} = 14$$

     |4   -6|   |9   -3|   |4  2|
I2 = |      | + |      | + |    |
     |-6  9 |   |-3  1 |   |2  1|

$$I_{3} = \left|\begin{matrix}4 & -6 & 2\\-6 & 9 & -3\\2 & -3 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}4 & -6 & 2 & 2\\-6 & 9 & -3 & -3\\2 & -3 & 1 & 1\\2 & -3 & 1 & -5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & -6 & 2\\-6 & 9 - \lambda & -3\\2 & -3 & 1 - \lambda\end{matrix}\right|$$

     |4  2 |   |9   -3|   |1  1 |
K2 = |     | + |      | + |     |
     |2  -5|   |-3  -5|   |1  -5|

     |4   -6  2 |   |9   -3  -3|   |4  2  2 |
     |          |   |          |   |        |
K3 = |-6  9   -3| + |-3  1   1 | + |2  1  1 |
     |          |   |          |   |        |
     |2   -3  -5|   |-3  1   -5|   |2  1  -5|

$$I_{1} = 14$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 14 \lambda^{2}$$
$$K_{2} = -84$$
$$K_{3} = 0$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge I_{4} = 0 \wedge K_{3} = 0 \wedge I_{1} \neq 0$$
then by type of surface:
you need to
then the canonical form of the equation will be
$$I_{1} \tilde x^{2} + \frac{K_{2}}{I_{1}} = 0$$
$$14 \tilde x^{2} - 6 = 0$$
$$\tilde x^{2} = \frac{3}{7}$$
this equation is fora type two parallel planes
- reduced to canonical form