The equations $x+y+z=3$, $x+2y+2z=6$ and $x+ay+3z=b$ have

Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 0 & 5 \\ 0 & 2 & 1 \end{bmatrix}$ and $b = \begin{bmatrix} 0 \\ -3 \\ 1 \end{bmatrix}$. Which of the following is true?

If the point $P(\alpha, \beta, \gamma)$ lies on the plane $2x + y + z = 1$ and $\begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} \begin{bmatrix} 1 & 9 & 1 \\ 8 & 2 & 1 \\ 7 & 3 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$,then $\alpha^2 + \beta^2 + \gamma^2 = $

Let $A = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix}$,where $i = \sqrt{-1}$. Then,the system of linear equations $A^{8} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 64 \end{bmatrix}$ has :

Let $S$ be the set of all real values of $k$ for which the system of linear equations $x + y + z = 2$,$2x + y - z = 3$,and $3x + 2y + kz = 4$ has a unique solution. Then $S$ is

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