If $\left| {\begin{array}{*{20}{c}}a&b&{a + b}\\b&c&{b + c}\\{a + b}&{b + c}&0\end{array}} \right| = 0$,then $a, b, c$ are in:

Solve the equation $\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$,where $a \neq 0$.

Let $\Delta = \left| \begin{array}{ccc} 1 & \omega & 2\omega^2 \\ 2 & 2\omega^2 & 4\omega^3 \\ 3 & 3\omega^3 & 6\omega^4 \end{array} \right|$ where $\omega$ is the cube root of unity,then

If $\alpha+\beta+\gamma=2 \pi$,then the system of equations
$x+(\cos \gamma) y+(\cos \beta) z=0$
$(\cos \gamma) x+y+(\cos \alpha) z=0$
$(\cos \beta) x+(\cos \alpha) y+z=0$
has :

The sum of all possible values of $\theta \in [0, 2\pi]$, for which the system of equations : $x \cos 3\theta - 8y - 12z = 0, x \cos 2\theta + 3y + 3z = 0, x + y + 3z = 0$ has a non-trivial solution, is equal to :

For real numbers $x, y$ and $z$,if $x \neq y \neq z$,$\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$ and $\left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{array}\right| \neq 0$,then $xyz = $ . . . . . . .