The system of equations $x+y+z=6$,$x+2y+5z=9$,$x+5y+\lambda z=\mu$ has no solution if

If the system of equations $x+2y-3z=2$,$2x+\lambda y+5z=5$,$14x+3y+\mu z=33$ has infinitely many solutions,then $\lambda+\mu$ is equal to:

If the system of linear equations $2x - 3y = \gamma + 5$ and $\alpha x + 5y = \beta + 1$,where $\alpha, \beta, \gamma \in R$,has infinitely many solutions,then the value of $|9\alpha + 3\beta + 5\gamma|$ is equal to

For the system $x-y+z=4, 2x+y-3z=0, x+y+z=2$,the values of $x, y, z$ respectively are given by

If $\begin{bmatrix} 2 & 1 & 1 \\ 0 & 3 & -1 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$, then $\begin{bmatrix} x \\ y \\ z \end{bmatrix} =$

Consider the simultaneous linear equations $\beta x + \alpha y - z = -1$, $3x - \beta y + \alpha z = 0$, and $\alpha x + \beta y + z = 1$. In the usual notation used in Cramer's rule, given that $\frac{\Delta_1}{\Delta} = -1$, $\frac{\Delta_2}{\Delta} = 1$, and $\frac{\Delta_3}{\Delta} = 2$, then $(\alpha, \beta) = $