-6x1x2+3x1x3 canonical form

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

     |0  0|   |0   -3|   | 0   3/2|
I2 = |    | + |      | + |        |
     |0  0|   |-3  0 |   |3/2   0 |

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

     |0  0|   |0  0|   |0  0|
K2 = |    | + |    | + |    |
     |0  0|   |0  0|   |0  0|

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

$$I_{1} = 0$$
$$I_{2} = - \frac{45}{4}$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{45 \lambda}{4}$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - \frac{45 \lambda}{4} = 0$$
$$\lambda_{1} = - \frac{3 \sqrt{5}}{2}$$
$$\lambda_{2} = \frac{3 \sqrt{5}}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$\frac{3 \sqrt{5} \tilde x2^{2}}{2} - \frac{3 \sqrt{5} \tilde x3^{2}}{2} = 0$$
$$- \frac{\tilde x2^{2}}{\left(\frac{5^{\frac{3}{4}} \sqrt{6}}{15}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{5^{\frac{3}{4}} \sqrt{6}}{15}\right)^{2}} = 0$$
this equation is fora type two intersecting planes
- reduced to canonical form