Let $$X = \left\{(x, y) \in \mathbb{Z} \times \mathbb{Z} : \frac{x^2}{8} + \frac{y^2}{20} < 1 \text{ and } y^2 < 5x\right\}$$. Three distinct points 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

Both inequalities restrict $$x, y$$ to small integers.
Ellipse: $$\dfrac{x^{2}}{8} + \dfrac{y^{2}}{20} \lt 1$$ gives $$y^{2} \lt 20$$, so $$y \in \{-4,-3,\dots,4\}$$.
Parabola: $$y^{2} \lt 5x$$ forces $$x \gt 0$$ and, because $$y^{2} \le 16$$, at most $$x = 1, 2$$ are possible (the ellipse allows $$|x| \lt \sqrt{8}\approx 2.83$$).

Checking every admissible $$y$$:

$$\begin{array}{c|c|c} y & \text{Ellipse bound on }|x| & \text{Parabola bound on }x \\ \hline \pm 4 & |x| \lt 1.264 & x \gt 3.2 & \text{No integer }x \\ \pm 3 & |x| \lt 2.098 & x \ge 2 & (2,\pm 3) \\ \pm 2 & |x| \lt 2.529 & x \ge 1 & (1,\pm 2),(2,\pm 2) \\ \pm 1 & |x| \lt 2.758 & x \ge 1 & (1,\pm 1),(2,\pm 1) \\ 0 & |x| \lt 2.828 & x \ge 1 & (1,0),(2,0) \end{array}$$

Thus the set $$X$$ contains exactly 12 lattice points:

$$(2,3),(2,-3);\\ (1,2),(2,2),(1,-2),(2,-2);\\ (1,1),(2,1),(1,-1),(2,-1);\\ (1,0),(2,0).$$

Total ways to choose three distinct points:
$$N_{\text{total}} = {}^{12}C_{3} = 220.$$ This matches the denominators given in the options.

Let us classify the points by the parity of their coordinates
$$(x \bmod 2,\; y \bmod 2):$$

EE (0,0): (2,2),(2,-2),(2,0) ⇒ 3 points
EO (0,1): (2,3),(2,-3),(2,1),(2,-1) ⇒ 4 points
OE (1,0): (1,2),(1,-2),(1,0) ⇒ 3 points
OO (1,1): (1,1),(1,-1) ⇒ 2 points

For any three lattice points the doubled area is
$$\Delta = x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2}).$$ Since $$x_i,y_i$$ are integers, $$\Delta$$ is an integer and the (actual) area is $$\dfrac{|\Delta|}{2}$$.

Parity test:
If at least two vertices have identical parity class, the vector joining them has both components even; hence every term in $$\Delta$$ is even and $$\Delta$$ itself is even. If the three vertices lie in three different parity classes, one easily verifies (substituting any representatives) that $$\Delta$$ is odd. Therefore

Area is an integer ⇔ not all three parity classes are different.

Counting triples with three different parity classes:

$$\begin{aligned} \text{EE, EO, OE:}&\; 3 \times 4 \times 3 = 36\\ \text{EE, EO, OO:}&\; 3 \times 4 \times 2 = 24\\ \text{EE, OE, OO:}&\; 3 \times 3 \times 2 = 18\\ \text{EO, OE, OO:}&\; 4 \times 3 \times 2 = 24\\ \hline \text{Total odd } \Delta &= 36+24+18+24 = 102 \end{aligned}$$

Thus triples with even $$\Delta$$ (integer area or zero area): $$N_{\text{even}} = 220 - 102 = 118.$$ We must now discard the degenerate cases (area $$=0$$).

Because all points have $$x = 1$$ or $$x = 2$$, the only straight lines containing three or more of them are the vertical lines $$x = 1$$ and $$x = 2$$:

x = 1 has 5 points ⇒ $${}^{5}C_{3}=10$$ collinear triples.
x = 2 has 7 points ⇒ $${}^{7}C_{3}=35$$ collinear triples.

Total degenerate triples $$N_{\text{col}} = 10 + 35 = 45.$$

Favourable triples (non-collinear with integer area):
$$N_{\text{fav}} = N_{\text{even}} - N_{\text{col}} = 118 - 45 = 73.$$

Probability required:
$$P = \dfrac{N_{\text{fav}}}{N_{\text{total}}} = \dfrac{73}{220}.$$

Option B which is: $$\dfrac{73}{220}$$