Find the area of the triangle with vertices at the points $(1,0), (6,0), (4,3)$.

If $\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right|$,then the two triangles with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and $(a_1, b_1), (a_2, b_2), (a_3, b_3)$ must be:

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 = $ . . . . . . .

Let $A=\begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix}$. If $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(3A)))=2^m \cdot 3^n$,where $m, n \in N$,then $m+n$ is equal to:

If $1, \omega, \omega^2$ are the cube roots of unity,then $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix} = $

The value of $\left|\begin{array}{lll}\sin ^2 14^{\circ} & \sin ^2 66^{\circ} & \tan 135^{\circ} \\ \sin ^2 66^{\circ} & \tan 135^{\circ} & \sin ^2 14^{\circ} \\ \tan 135^{\circ} & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right|$ is