$(x_1,y_1),(x_2,y_2)$ & $(x_3,y_3)$ are the vertices which lie on the parabola hence

$x_1=\large\frac{y_1^2}{4a}$

$x_2=\large\frac{y_2^2}{4a}$

$x_3=\large\frac{y_3^2}{4a}$

Now area of triangle with $(\large\frac{y_1^2}{4a}$$,y_1),(\large\frac{y_2^2}{4a}$$,y_2),(\large\frac{y_3^2}{4a}$$,y_3)$ is

$\large\frac{1}{2}$$\begin{vmatrix}\large\frac{y_1^2}{4a}&y_1&1\\\large\frac{y_2^2}{4a}&y_2&1\\\large\frac{y_3^2}{4a}&y_3&1\end{vmatrix}$

Hence area of triangle=$\large\frac{1}{2}\mid\frac{y_1^2}{4a}$$(y_2-y_3)-y_1(\large\frac{y_2^2}{4a}-\frac{y_3^2}{4a})$$+1(\large\frac{y_2^2y_3}{4a}-\frac{y_2y_3^2}{4a})\mid$

$\Rightarrow \large\frac{1}{2}\mid \big(\large\frac{y_1^2y_2-y_1^2y_3-y_1y_2^2+y_1y_3^2+y_2^2y_3-y_2y_3^2}{4a}\big)\mid$

$\Rightarrow \large\frac{1}{8a}$$\mid(y_1-y_2)(y_2-y_3)(y_3-y_1)\mid$

Hence (a) is the correct answer.