$A$ and $C$ lie in $\left[0, \frac{\pi}{2}\right)$ and $B$ lies in $[0, 2\pi]$. If $\tan A + 3 \cos B + 6 \sin C = 1$; $3 \tan A + \cos B + 4 \sin C = 4$; $5 \tan A + 3 \cos B - 8 \sin C = -2$, then $B - 2A - C =$

If the system of linear equations $2x + 2y + 3z = a$,$3x - y + 5z = b$,and $x - 3y + 2z = c$,where $a, b, c$ are non-zero real numbers,has more than one solution,then:

For a real number $\alpha$,if the system of linear equations $\begin{bmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$ has infinitely many solutions,then $1+\alpha+\alpha^2=$

Solve the system of linear equations using the matrix method: $2x + y + z = 1$,$x - 2y - z = \frac{3}{2}$,and $3y - 5z = 9$.

Solve the system of linear equations using the matrix method: $2x + 3y + 3z = 5$,$x - 2y + z = -4$,$3x - y - 2z = 3$.

The number of solutions of the equations $x + 4y - z = 0,$ $3x - 4y - z = 0,$ and $x - 3y + z = 0$ is