Explanation

(c)
Let the points are $\left( x_{1},\ y_{1} \right),\left( x_{2},y_{2} \right)$ and $\left( x_{3},\ y_{3} \right)$
$\therefore y_{1}^{2} = {4ax}_{1},\ y_{2}^{2} = {4ax}_{2},\ y_{3}^{2} = {4ax}_{3}$
$\therefore$ Area of triangle whose vertices are $\left( x_{1},\ y_{1} \right),\left( x_{2},y_{2} \right)$ and $\left( x_{3},\ y_{3} \right)$
$= \frac{1}{2}\left\| \begin{matrix}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{matrix} \right\| = \frac{1}{2}\left\| \begin{matrix}
\frac{y_{1}^{2}}{4a} & y_{1} & 1 \\
\frac{y_{2}^{2}}{4a} & y_{2} & 1 \\
\frac{y_{3}^{2}}{4a} & y_{3} & 1
\end{matrix} \right\| = \frac{1}{8a}\left\| \begin{matrix}
y_{1}^{2} & y_{1} & 1 \\
y_{2}^{2} & y_{2} & 1 \\
y_{3}^{2} & y_{3} & 1
\end{matrix} \right\|$
$\Rightarrow$ Area of triangle
$= \frac{1}{8a}\left( y_{1} - y_{2} \right)\left( y_{2} - y_{3} \right)\left( y_{3} - y_{1} \right)$