If $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$,then $\det A$ =

If $k > 1$ and the determinant of the matrix $A^2$, where $A = \begin{bmatrix} k & k\alpha & \alpha \\ 0 & \alpha & k\alpha \\ 0 & 0 & k \end{bmatrix}$, is $k^2$, then $|\alpha|$ is equal to

If $A = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix}$,$B = \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}$,and $C = \begin{vmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}$,then which relation is correct?

If $\omega$ is a cube root of unity,then the root of the equation $\left| \begin{array}{ccc} x + 2 & \omega & \omega^2 \\ \omega & x + 1 + \omega^2 & 1 \\ \omega^2 & 1 & x + 1 + \omega \end{array} \right| = 0$ is:

Let $A(a, 0)$,$B(b, 2b+1)$,and $C(0, b)$,where $b \neq 0$ and $|b| \neq 1$,be points such that the area of triangle $ABC$ is $1 \, \text{sq. unit}$. Then,the sum of all possible values of $a$ is:

If the matrix $\begin{bmatrix} 1 & 3 & \lambda + 2 \\ 2 & 4 & 8 \\ 3 & 5 & 10 \end{bmatrix}$ is singular,then $\lambda = $