The system of equations $3x + 2y + z = 6$, $3x + 4y + 3z = 14$ and $6x + 10y + 8z = a$ has an infinite number of solutions if $a$ is equal to

For the system of equations $x+y+z=6$,$x+2y+\alpha z=10$,and $x+3y+5z=\beta$,which one of the following is $NOT$ true?

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

If $\begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & -5 \\ 1 & 2 & 5 \end{bmatrix} = \begin{bmatrix} 3 & 5 & 2 \end{bmatrix}$, then $\alpha^3 + \beta^3 + \gamma^3 = $

Two fair dice are thrown. The numbers on them are taken as $\lambda$ and $\mu$,and a system of linear equations
$x+y+z=5$
$x+2y+3z=\mu$
$x+3y+\lambda z=1$
is constructed. If $p$ is the probability that the system has a unique solution and $q$ is the probability that the system has no solution,then:

For how many values of $k$ does the system of linear equations $(k + 1)x + 8y = 4k$ and $kx + (k + 3)y = 3k - 1$ have no solutions?