If the system of equations $x + y - z = 0, 3x - \alpha y - 3z = 0, x - 3y + z = 0$ has a non-zero solution,then $\alpha = $

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

Let $S$ be the set of values of $\lambda$,for which the system of equations
$6 \lambda x - 3 y + 3 z = 4 \lambda^2$
$2 x + 6 \lambda y + 4 z = 1$
$3 x + 2 y + 3 \lambda z = \lambda$
has no solution. Then $12 \sum_{\lambda \in S} |\lambda|$ is equal to $...........$.

The set of real values of $\alpha$ for which the system of linear equations
$\begin{aligned}
& x+(\sin \alpha) y+(\cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& -x+(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned}$
has a non-trivial solution is

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 $1+\alpha+\alpha^2=$

The product of all real values of $b$ such that there is no solution to the system of equations $2x + 5y + z = 19$,$-4x + by + 6z = -42$,and $-3y - bz = 81$ is: