If $(a, b), (c, d)$ and $(e, f)$ are the vertices of $\Delta ABC$ and $\Delta$ denotes the area of $\Delta ABC$, then $\begin{vmatrix} a & c & e \\ b & d & f \\ 1 & 1 & 1 \end{vmatrix}^2$ is equal to:

$4\Delta^2$

The correct answer is Option (2) → $4\Delta^2$ ##

$\Delta = \frac{1}{2} \begin{vmatrix} a & c & e \\ b & d & f \\ 1 & 1 & 1 \end{vmatrix}$

$2\Delta = \begin{vmatrix} a & c & e \\ b & d & f \\ 1 & 1 & 1 \end{vmatrix}$

$4\Delta^2 = \begin{vmatrix} a & c & e \\ b & d & f \\ 1 & 1 & 1 \end{vmatrix}^2$