If the vertices of a triangle are $(-2, 6)$,$(3, -6)$,and $(1, 5)$,then the area of the triangle is

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?

Let $P$ be a matrix of order $3 \times 3$ such that all the entries in $P$ are from the set $\{-1, 0, 1\}$. Then,the maximum possible value of the determinant of $P$ is:

The value of $\left|\begin{array}{lll}x & p & q \\ p & x & q \\ p & q & x\end{array}\right|$ is

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:

The area of the triangle with vertices $(3, 8)$, $(-4, 2)$ and $(5, 1)$ is $\frac{P}{4}$. The value of $P$ is: