Let $\mathrm{X}=\left\{(\mathrm{x},\mathrm{y})\in \mathrm{\mathbb{Z}}\times \mathrm{\mathbb{Z}}:\frac{{\mathrm{x}}^2}{8}+\frac{{\mathrm{y}}^2}{20}<1\text{ and }{\mathrm{y}}^2<5\mathrm{x}\right\}$. Three distinct point P, Q and R are randomly chosen from X. Then the probability that P, Q and R form a triangle whose area is a positive integer, is

$\begin{array}{l}\frac{{\mathrm{x}}^2}{8}+\frac{{\mathrm{y}}^2}{20}<1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{y}}^2<5\mathrm{x}\\ \frac{{\mathrm{x}}^2}{8}+\frac{5\mathrm{x}}{20}=1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\Rightarrow \frac{{\mathrm{x}}^2}{8}+\frac{\mathrm{x}}{4}=1\Rightarrow {\mathrm{x}}^2+2\mathrm{x}=8\\ \Rightarrow (\mathrm{x}+1{)}^2=9\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\Rightarrow \mathrm{x}=2\\ \mathrm{x}=1,{\mathrm{y}}^2<5\Rightarrow \mathrm{y}<\sqrt{5}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ (1,0)(1,1)(1,2)(1,-1)(1,-2)\\ (2,0)(2,1)(2,2)(2,3)(2,-1)(2,-2)(2,-3)\end{array}$
Number of ways to select 3 points ${=}^{12}{\mathrm{C}}_3$
$=\frac{12\times 11\times 10}{6}=2\times 11\times 10=220$
Area of  $∆$PQR is an integer 
Triangle has two vertices on x = 1 or x = 2 
$\left(1,{\mathrm{y}}_1\right)\left(1,{\mathrm{y}}_2\right)\left(2,{\mathrm{y}}_3\right)\text{ (or) }\left(1,{\mathrm{y}}_3\right)\left(2,{\mathrm{y}}_1\right)\left(2,{\mathrm{y}}_2\right)$
Area of $\mathrm{\Delta \ell e}=\frac{1}{2}\left|\left(1\right)\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right|$
$\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)$  is an even integer 
Number of ways = ${\left({ }^3{\mathrm{C}}_2{+}^2{\mathrm{C}}_2\right)}^7{\mathrm{C}}_1$ for x = 1
$={\left({ }^4{\mathrm{C}}_2{+}^3{\mathrm{C}}_2\right)}^5{\mathrm{C}}_1\text{ for }\mathrm{x}=2$
Number of triangles = 28 + 45 = 73 
Probability = $\frac{73}{220}$