If the system of linear equations
$7x + 11y + \alpha z = 13$
$5x + 4y + 7z = \beta$
$175x + 194y + 57z = 361$
has infinitely many solutions,then $\alpha + \beta + 2$ is equal to

If $A=\begin{bmatrix} x & y & y \\ y & x & y \\ y & y & x \end{bmatrix}$ is a matrix such that $5 A^{-1}=\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}$, then $A^2-4 A=$

Let the system of linear equations $x+y+kz=2$; $2x+3y-z=1$; $3x+4y+2z=k$ have infinitely many solutions. Then the system $(k+1)x+(2k-1)y=7$; $(2k+1)x+(k+5)y=10$ has:

Let $[\lambda]$ be the greatest integer less than or equal to $\lambda$. The set of all values of $\lambda$ for which the system of linear equations $x+y+z=4$,$3x+2y+5z=3$,$9x+4y+(28+[\lambda])z=[\lambda]$ has a solution is:

Consider the following system of equations in matrix form $\begin{bmatrix} 1 \\ 2 \\ \lambda \end{bmatrix} \begin{bmatrix} 1 & 2 & \lambda \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0$. Then which one of the following statements is true?

If the values $x=\alpha, y=\beta, z=\gamma$ satisfy all the $3$ equations $x+2y+3z=4$,$3x+y+z=3$ and $x+3y+3z=2$,then $3\alpha+\gamma=$