x1^2+x2^2-2x3^2-14x1x2+2sqrt(6)x1x3+2sqrt(6)x2x3 canonical form

Given equation of the surface of 2-order:
$$x_{1}^{2} - 14 x_{1} x_{2} + 2 \sqrt{6} x_{1} x_{3} + x_{2}^{2} + 2 \sqrt{6} x_{2} x_{3} - 2 x_{3}^{2} = 0$$
This equation looks like:
$$a_{11} x_{3}^{2} + 2 a_{12} x_{2} x_{3} + 2 a_{13} x_{1} x_{3} + 2 a_{14} x_{3} + a_{22} x_{2}^{2} + 2 a_{23} x_{1} x_{2} + 2 a_{24} x_{2} + a_{33} x_{1}^{2} + 2 a_{34} x_{1} + a_{44} = 0$$
where
$$a_{11} = -2$$
$$a_{12} = \sqrt{6}$$
$$a_{13} = \sqrt{6}$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = -7$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 0$$

     |         ___|              |         ___|
     | -2    \/ 6 |   |1   -7|   | -2    \/ 6 |
I2 = |            | + |      | + |            |
     |  ___       |   |-7  1 |   |  ___       |
     |\/ 6     1  |              |\/ 6     1  |

$$I_{3} = \left|\begin{matrix}-2 & \sqrt{6} & \sqrt{6}\\\sqrt{6} & 1 & -7\\\sqrt{6} & -7 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}-2 & \sqrt{6} & \sqrt{6} & 0\\\sqrt{6} & 1 & -7 & 0\\\sqrt{6} & -7 & 1 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 2 & \sqrt{6} & \sqrt{6}\\\sqrt{6} & 1 - \lambda & -7\\\sqrt{6} & -7 & 1 - \lambda\end{matrix}\right|$$

     |-2  0|   |1  0|   |1  0|
K2 = |     | + |    | + |    |
     |0   0|   |0  0|   |0  0|

     |         ___   |                 |         ___   |
     | -2    \/ 6   0|   |1   -7  0|   | -2    \/ 6   0|
     |               |   |         |   |               |
K3 = |  ___          | + |-7  1   0| + |  ___          |
     |\/ 6     1    0|   |         |   |\/ 6     1    0|
     |               |   |0   0   0|   |               |
     |  0      0    0|                 |  0      0    0|

$$I_{1} = 0$$
$$I_{2} = -64$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 64 \lambda$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - 64 \lambda = 0$$
$$\lambda_{1} = -8$$
$$\lambda_{2} = 8$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$8 \tilde x2^{2} - 8 \tilde x3^{2} = 0$$
$$- \frac{\tilde x2^{2}}{\left(\frac{\sqrt{2}}{4}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{\sqrt{2}}{4}\right)^{2}} = 0$$
this equation is fora type two intersecting planes
- reduced to canonical form