The area of a triangle whose vertices are $(1, -1)$,$(-1, 1)$,and $(-1, -1)$ is given by

If $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ are the vertices of a triangle whose area is $k$ square units,then $\left|\begin{array}{ccc}x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4\end{array}\right|^{2}$ is (in $k^{2}$)

Evaluate the determinant: $\left| \begin{array}{ccc} 1 & a & b \\ -a & 1 & c \\ -b & -c & 1 \end{array} \right|$

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 = $ . . . . . . .

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:

Let $\left| {\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}} \right| = x + iy$,then