If the system of linear equations,$x+y+z = 6$,$x+2y+3z = 10$,and $3x+2y+\lambda z = \mu$ has more than two solutions,then $\mu-\lambda^{2}$ is equal to

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 number of values of $k$ for which the system of linear equations,$(k + 2)x + 10y = k$ and $kx + (k + 3)y = k - 1$ has no solution,is

If $\left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}2 \\ 4\end{array}\right]$,then the values of $x$ and $y$ respectively are

Let $\beta$ be a real number. Consider the matrix $A = \begin{bmatrix} \beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2 \end{bmatrix}$. If $A^7 - (\beta - 1)A^6 - \beta A^5$ is a singular matrix,then the value of $9\beta$ is:

The sum of three numbers is $6$. Thrice the third number when added to the first number gives $7$. On adding three times the first number to the sum of the second and third numbers,we get $12$. The product of these numbers is: