Area of $\Delta ABC$ is $\frac{3}{2}$. The $Y$-coordinate of its centroid is $8$ less than three times its $X$-coordinate. If $A$ is $(2, -3)$ and $B$ is $(3, -2)$,find the coordinates of $C$.

(A) Let the coordinates of $C$ be $(x, y)$.
The coordinates of the centroid $G$ of $\Delta ABC$ are given by $\left(\frac{2+3+x}{3}, \frac{-3-2+y}{3}\right) = \left(\frac{x+5}{3}, \frac{y-5}{3}\right)$.
Given that the $Y$-coordinate of the centroid is $8$ less than three times its $X$-coordinate:
$\frac{y-5}{3} = 3\left(\frac{x+5}{3}\right) - 8$
$\frac{y-5}{3} = x + 5 - 8$
$\frac{y-5}{3} = x - 3$
$y - 5 = 3x - 9 \implies 3x - y = 4$ ... $(1)$
The area of $\Delta ABC$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = \frac{3}{2}$.
$\frac{1}{2} |2(-2 - y) + 3(y - (-3)) + x(-3 - (-2))| = \frac{3}{2}$
$|-4 - 2y + 3y + 9 - x| = 3$
$|y - x + 5| = 3$
This gives two cases: $y - x + 5 = 3$ or $y - x + 5 = -3$.
Case $1$: $y - x = -2 \implies x - y = 2$ ... $(2)$
Case $2$: $y - x = -8 \implies x - y = 8$ ... $(3)$
Solving $(1)$ and $(2)$: $3x - y = 4$ and $x - y = 2$. Subtracting gives $2x = 2 \implies x = 1$. Then $y = -1$.
Solving $(1)$ and $(3)$: $3x - y = 4$ and $x - y = 8$. Subtracting gives $2x = -4 \implies x = -2$. Then $y = -10$.
Thus,the coordinates of $C$ are $(1, -1)$ or $(-2, -10)$.