If $a > 0$and discriminant of $a{x^2} + 2bx + c$is negative, then $\left| {\,\begin{array}{*{20}{c}}a&b&{ax + b}\\b&c&{bx + c}\\{ax + b}&{bx + c}&0\end{array}\,} \right|$ is

Find area of the triangle with vertices at the point given in each of the following: $(1,0),(6,0),(4,3)$

Let $S$ be the set of all values of $\theta \in[-\pi, \pi]$ for which the system of linear equations

$x+y+\sqrt{3} z=0$

$-x+(\tan \theta) y+\sqrt{7} z=0$

$x+y+(\tan \theta) z=0$

has non-trivial solution. Then $\frac{120}{\pi} \sum_{\theta \in s} \theta$ is equal to

$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2} - bc}\\1&b&{{b^2} - ac}\\1&c&{{c^2} - ab}\end{array}\,} \right| = $

If $\left| {\begin{array}{*{20}{c}}{x - 4}&{2x}&{2x}\\{2x}&{x - 4}&{2x}\\{2x}&{2x}&{x - 4}\end{array}} \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2},$ then the ordered pair $\left( {A,B} \right) = $. . . . .

If $a,b,c$ and $d $ are complex numbers, then the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}2&{a + b + c + d}&{ab + cd}\\{a + b + c + d}&{2(a + b)(c + d)}&{ab(c + d) + cd(a + b)}\\{ab + cd}&{ab(c + d) + cd(a + d)}&{2abcd}\end{array}} \right|$is