The number of solutions of the system of equations $2x + y - z = 7$,$x - 3y + 2z = 1$,and $x + 4y - 3z = 5$ is:

Consider the system of equations
$\begin{cases} x+y+z = 0 \\ \alpha x+\beta y+\gamma z = 0 \\ \alpha^{2} x+\beta^{2} y+\gamma^{2} z = 0 \end{cases}$
Then the system of equations has

If the solution of the system of simultaneous linear equations $x+y-z=6$,$3x+2y-z=5$ and $2x-y-2z+3=0$ is $x=\alpha, y=\beta, z=\gamma$,then $\alpha+\beta=$

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:

If $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$,then $A^{2} + xA + yI = 0$ for $(x, y)$ is

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