Area of the triangle ABC with vertices $\left(a, a^2\right),\left(b, b^2\right)$ and $\left(c, c^2\right)$ is equal to:

$\frac{1}{2}|a b c(a-b)(b-c)(c-a)|$ $\frac{1}{2}\left|\frac{1}{a b c}(a-b)(b-c)(c-a)\right|$ $\frac{1}{2}|(a-b)(b-c)(c-a)|$ $\frac{1}{2}[(a-b)(b-c)(c-a)]^2$

$\frac{1}{2}|(a-b)(b-c)(c-a)|$

The correct answer is Option (3) → $\frac{1}{2}|(a-b)(b-c)(c-a)|$