For $\alpha, \beta \in R$, suppose the system of linear equations $x-y+z=5$ ; $ 2 x+2 y+\alpha z=8 $ ; $3 x-y+4 z=\beta$ has infinitely many solutions. Then $\alpha$ and $\beta$ are the roots of

If the system of linear equations $2 x+3 y-z=-2$  ; $x+y+z=4$  ; $x-y+|\lambda| z=4 \lambda-4$  (where $\lambda \in R$), has no solution, then

$\left| {\,\begin{array}{*{20}{c}}{1 + i}&{1 - i}&i\\{1 - i}&i&{1 + i}\\i&{1 + i}&{1 - i}\end{array}\,} \right| = $

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

The values of $\alpha$, for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$, lie in the interval

If $a > 0$and discriminant of $a{x^2} + 2bx + c$is negative, then $\left| {\,\begin{array}{*{20}{c}}a&b&{ax + b}\\b&c&{bx + c}\\{ax + b}&{bx + c}&0\end{array}\,} \right|$ is