If $A (x_1,y_1), B(x_2, y_2)$ and $C (x_3, y_3)$ are vertices of an equilateral triangle whose each side is equal to a, then find $\begin{vmatrix}x_1&y_1&2\\x_2& y_2&2\\x_3&y_3&2\end{vmatrix}$ is equal to

$3a^4$

Let Δ be the area of triangle ABC. Then,

$Δ=\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2& y_2&1\\x_3&y_3&1\end{vmatrix}⇒2Δ\begin{vmatrix}x_1&y_1&1\\x_2& y_2&1\\x_3&y_3&1\end{vmatrix}$

$⇒4Δ=2\begin{vmatrix}x_1&y_1&1\\x_2& y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}x_1&y_1&2\\x_2& y_2&2\\x_3&y_3&2\end{vmatrix}$

$⇒16Δ^2=\begin{vmatrix}x_1&y_1&2\\x_2& y_2&2\\x_3&y_3&2\end{vmatrix}^2$  ...(i)

But, the area of equilateral triangle with each side equal to a is $\frac{\sqrt{3}}{4}a^2$.

$∴Δ=\frac{\sqrt{3}}{4}a^2⇒16Δ^2=3a^4$   ...(ii)

Form (i) and (ii), we obtain

$\begin{vmatrix}x_1&y_1&2\\x_2& y_2&2\\x_3&y_3&2\end{vmatrix}^2=3a^4$