If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^3| = 27$,then $\alpha = $

If $\left| \begin{matrix} 1 & a & a^2 \\ 1 & x & x^2 \\ b^2 & ab & a^2 \end{matrix} \right| = 0$,then:

If $a, b, c$ are in $A.P.$,then the determinant $\left|\begin{array}{lll}x+2 & x+3 & x+2a \\ x+3 & x+4 & x+2b \\ x+4 & x+5 & x+2c\end{array}\right|$ is

If $\left|\begin{array}{lll}a & a^3 & a^4 \\ b & b^3 & b^4 \\ c & c^3 & c^4\end{array}\right|=k(a-b)(b-c)(c-a)$ then $k=$

If $n \ne 3k$ and $1, \omega, \omega^2$ are the cube roots of unity,then $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^n \\ \omega^n & \omega^{2n} & 1 \end{vmatrix}$ has the value

If $\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1+y & 1+2 y & 1 \\ 1+z & 1+z & 1+3 z\end{array}\right| = 10 k x y z \left(3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$,then $k = \text{ . . . . . . }$ (where $x, y, z \neq 0$ and $3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \neq 0$).