If the coordinates of the vertices of an equilateral triangle with sides of length '$a$' are $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$ and $\left( {{x_3},{y_3}} \right)$, then
${\left| {\begin{array}{llllllllllllllllllll}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right|^2} = \frac{{3{a^4}}}{4}$

Since, we know that area of a triangle with vertices $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$ and $\left( {{x_3},{y_3}} \right)$, is given by

$\Delta = \frac{1}{2}\left| {\begin{array}{llllllllllllllllllll}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right|$

$\Rightarrow$ ${\Delta ^2} = \frac{1}{4}{\left| {\begin{array}{llllllllllllllllllll}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right|^2}$

……..(i)
We know that, area of an equilateral triangle with side $a$, $\Delta = \frac{1}{2}\left( {\frac{{\sqrt 3 }}{2}} \right){a^2} = \frac{{\sqrt 3 }}{4}{a^2}$
$\Rightarrow$ ${\Delta ^2} = \frac{3}{{16}}{a^4}$

…….(ii)

From Eqs. (i) and (ii),
$\frac{3}{{16}}{a^4} = \frac{1}{4}{\left| {\begin{array}{llllllllllllllllllll}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right|^2}$

$\Rightarrow$ ${\left| {\begin{array}{llllllllllllllllllll}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right|^2} = \frac{3}{4}{a^4}$