Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants and find $k$ if $D(k, 0)$ is a point such that the area of triangle $ABD$ is $3 \, \text{sq units}$.

If $a x^{4}+b x^{3}+c x^{2}+d x+e = \left|\begin{array}{ccc}x^{3}+3 x & x-1 & x+3 \\ x+1 & -2 x & x-4 \\ x-3 & x+4 & 3 x\end{array}\right|$,then $e$ is equal to

If $\omega$ is an imaginary root of unity,then the value of $\left| \begin{array}{ccc} a & b\omega^2 & a\omega \\ b\omega & c & b\omega^2 \\ c\omega^2 & a\omega & c \end{array} \right|$ is

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

Let $x, y, z > 0$ be the $2^{nd}, 3^{rd}, 4^{th}$ terms of a $G.P.$ respectively,and $\Delta = \begin{vmatrix} x^k & x^{k+1} & x^{k+2} \\ y^k & y^{k+1} & y^{k+2} \\ z^k & z^{k+1} & z^{k+2} \end{vmatrix} = (r-1)^2 \left(1 - \frac{1}{r^2}\right)$,where $r$ is the common ratio. Then $k = \dots$

The value of $a$ for which the system of equations $a^3x + (a + 1)^3y + (a + 2)^3z = 0$,$ax + (a + 1)y + (a + 2)z = 0$,and $x + y + z = 0$ has a non-zero solution is: