0x^2+5y^2-3z^2+14yz−2y+10z-3=0. canonical form

Given equation of the surface of 2-order:
$$5 y^{2} + 14 y z - 2 y - 3 z^{2} + 10 z - 3 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 5$$
$$a_{23} = 7$$
$$a_{24} = -1$$
$$a_{33} = -3$$
$$a_{34} = 5$$
$$a_{44} = -3$$
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
$$I_{1} = 2$$

     |0  0|   |5  7 |   |0  0 |
I2 = |    | + |     | + |     |
     |0  5|   |7  -3|   |0  -3|

$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 5 & 7\\0 & 7 & -3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & 0\\0 & 5 & 7 & -1\\0 & 7 & -3 & 5\\0 & -1 & 5 & -3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & 5 - \lambda & 7\\0 & 7 & - \lambda - 3\end{matrix}\right|$$

     |0  0 |   |5   -1|   |-3  5 |
K2 = |     | + |      | + |      |
     |0  -3|   |-1  -3|   |5   -3|

     |0  0   0 |   |5   7   -1|   |0  0   0 |
     |         |   |          |   |         |
K3 = |0  5   -1| + |7   -3  5 | + |0  -3  5 |
     |         |   |          |   |         |
     |0  -1  -3|   |-1  5   -3|   |0  5   -3|

$$I_{1} = 2$$
$$I_{2} = -64$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 2 \lambda^{2} + 64 \lambda$$
$$K_{2} = -32$$
$$K_{3} = 0$$
Because
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - 2 \lambda^{2} - 64 \lambda = 0$$
$$\lambda_{1} = 1 - \sqrt{65}$$
$$\lambda_{2} = 1 + \sqrt{65}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$\tilde x^{2} \left(1 - \sqrt{65}\right) + \tilde y^{2} \left(1 + \sqrt{65}\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{-1 + \sqrt{65}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{1 + \sqrt{65}}}\right)^{2}} = 0$$
this equation is fora type two intersecting planes
- reduced to canonical form