Given that,$a \alpha^2+2 b \alpha+c \neq 0$ and that the system of equations
$\begin{aligned} & (a \alpha+b) x+a y+b z=0 \\ & (b \alpha+c) x+b y+c z=0 \\ & (a \alpha+b) y+(b \alpha+c) z=0\end{aligned}$
has a non-trivial solution,then $a, b$ and $c$ lie in

What is the area of the triangle with vertices $(4, 4)$,$(3, -2)$,and $(3, -16)$?

Let $\alpha, \beta, \gamma$ be the real roots of the equation $x^{3} + ax^{2} + bx + c = 0$,where $a, b, c \in R$ and $a, b \neq 0$. If the system of equations in $u, v, w$ given by $\alpha u + \beta v + \gamma w = 0$,$\beta u + \gamma v + \alpha w = 0$,and $\gamma u + \alpha v + \beta w = 0$ has a non-trivial solution,then the value of $\frac{a^{2}}{b}$ is:

If $a_{1}, a_{2}, a_{3}, \ldots, a_{9}$ are in $AP$,then the value of $\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|$ is

If the coordinates of the points $A, B,$ and $C$ are $(4, 4), (3, -2),$ and $(3, -16)$ respectively,then the area of the triangle $ABC$ is

$\left| {\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}} \right| = $