4xx+yy+zz-4xy+4xz-3yz canonical form

Given equation of the surface of 2-order:
$$4 x^{2} - 4 x y + 4 x z + y^{2} - 3 y z + z^{2} = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -2$$
$$a_{13} = 2$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = - \frac{3}{2}$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 6$$

     |4   -2|   | 1    -3/2|   |4  2|
I2 = |      | + |          | + |    |
     |-2  1 |   |-3/2   1  |   |2  1|

$$I_{3} = \left|\begin{matrix}4 & -2 & 2\\-2 & 1 & - \frac{3}{2}\\2 & - \frac{3}{2} & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}4 & -2 & 2 & 0\\-2 & 1 & - \frac{3}{2} & 0\\2 & - \frac{3}{2} & 1 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & -2 & 2\\-2 & 1 - \lambda & - \frac{3}{2}\\2 & - \frac{3}{2} & 1 - \lambda\end{matrix}\right|$$

     |4  0|   |1  0|   |1  0|
K2 = |    | + |    | + |    |
     |0  0|   |0  0|   |0  0|

     |4   -2  0|   | 1    -3/2  0|   |4  2  0|
     |         |   |             |   |       |
K3 = |-2  1   0| + |-3/2   1    0| + |2  1  0|
     |         |   |             |   |       |
     |0   0   0|   | 0     0    0|   |0  0  0|

$$I_{1} = 6$$
$$I_{2} = - \frac{5}{4}$$
$$I_{3} = -1$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 6 \lambda^{2} + \frac{5 \lambda}{4} - 1$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Because

I3 != 0

then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
or
$$\lambda^{3} - 6 \lambda^{2} - \frac{5 \lambda}{4} + 1 = 0$$
$$\lambda_{1} = - \frac{1}{2}$$
$$\lambda_{2} = \frac{13}{4} - \frac{\sqrt{137}}{4}$$
$$\lambda_{3} = \frac{\sqrt{137}}{4} + \frac{13}{4}$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- \frac{\tilde x^{2}}{2} + \tilde y^{2} \left(\frac{13}{4} - \frac{\sqrt{137}}{4}\right) + \tilde z^{2} \left(\frac{\sqrt{137}}{4} + \frac{13}{4}\right) = 0$$
$$- \frac{\tilde x^{2}}{\left(\sqrt{2}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{13}{4} - \frac{\sqrt{137}}{4}}}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{1}{\sqrt{\frac{\sqrt{137}}{4} + \frac{13}{4}}}\right)^{2}}\right) = 0$$
this equation is fora type cone
- reduced to canonical form