If the system of equations $x + ay = 0,$ $az + y = 0$ and $ax + z = 0$ has infinite solutions, then the value of $a$ is

If the following system of linear equations

$2 x+y+z=5$

$x-y+z=3$

$x+y+a z=b$

has no solution, then :

Let $S$ be the set of all $\lambda \in \mathrm{R}$ for which the system of linear equations

$2 x-y+2 z=2$

$x-2 y+\lambda z=-4$

$x+\lambda y+z=4$

has no solution. Then the set $S$

For $\alpha, \beta \in \mathrm{R}$ and a natural number $\mathrm{n}$, let

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$. Then $2 A_{10}-A_8$

If the system of equation $3x - 2y + z = 0$, $\lambda x - 14y + 15z = 0$, $x + 2y + 3z = 0$ have a non-trivial solution, then $\lambda = $

If system of equations $kx + 2y - z = 2,$$\left( {k - 1} \right)x + ky + z = 1,x + \left( {k - 1} \right)y + kz = 3$ has only one solution, then number of possible real value$(s)$ of $k$ is -