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

The area of the triangle with vertices $(a, b)$,$(x_1, y_1)$,and $(x_2, y_2)$,where $a, x_1, x_2$ are in $G.P.$ with common ratio $r$ and $b, y_1, y_2$ are in $G.P.$ with common ratio $s$,is given by

An equilateral triangle has each of its sides of length $6 \text{ cm}$. If $(x_1, y_1), (x_2, y_2), \text{ and } (x_3, y_3)$ are its vertices,then the value of the determinant $\left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right|^2$ is equal to:

If $a, b, c$ are non-zero real numbers and if the equations $(a-1) x=y+z, (b-1) y=z+x, (c-1) z=x+y$ have a non-trivial solution,then $ab+bc+ca=$

If $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$,then the value of $x$ is . . . . . . .

The number of real values of $t$ such that the system of homogeneous equations
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
\end{aligned}$
has non-trivial solutions is