If the determinant of the matrix $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & \beta \\ -b & -\alpha & 0 \end{bmatrix}$ is zero for all $a, b$,then $\alpha + \beta =$

The area of the triangle whose vertices are $(3,5), (2,2)$ and $(k, 2)$ is $3$ sq. unit. Then,the value of $k$ is . . . . . . .

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $\mathbb{R},$ the determinant $\left| \begin{array}{ccc} y + 1 & \alpha & \beta \\ \alpha & y + \beta & 1 \\ \beta & 1 & y + \alpha \end{array} \right|$ is equal to

If $t_1, t_2$ and $t_3$ are distinct,then the points $(t_1, 2at_1 + at_1^3)$,$(t_2, 2at_2 + at_2^3)$ and $(t_3, 2at_3 + at_3^3)$ are collinear if:

If $a \neq b \neq c$,the value of $x$ which satisfies the equation $\left| \begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array} \right| = 0$ is

$\left| {\begin{array}{ccc} 19 & 17 & 15 \\ 9 & 8 & 7 \\ 1 & 1 & 1 \end{array}} \right| = $