If $\mathrm{y}_{1}, \mathrm{y}_{2}, \mathrm{y}_{3}$ are the ordinates of the vertices of a triangle inscribed in the parabola $\mathrm{y}^{2}=4 \mathrm{ax}(\mathrm{a}>0)$, then its area is
(a) $\frac{1}{8 \mathrm{a}}\left|\mathrm{y}_{1} \mathrm{y}_{2} \mathrm{y}_{3}\right|$
(b) $\frac{1}{8 \mathrm{a}}\left|\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)\right|$
(c) $\frac{1}{8 \mathrm{a}}\left|\left(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}\right)\right|$
(d) $8 a\left|\left(y_{1}-y_{2}\right)\left(y_{2}-y_{3}\right)\left(y_{3}-y_{1}\right)\right|$