Question:

<p><strong>Step 1:</strong> Let $\frac{s-a}{4} = \frac{s-b}{3} = \frac{s-c}{2} = k$ for some constant $k > 0$.</p><p>Then: $s-a = 4k$, $s-b = 3k$, $s-c = 2k$</p><p><strong>Step 2:</strong> From $2s = a+b+c$ and the above relations:</p><p>$s = (s-a) + (s-b) + (s-c) = 4k + 3k + 2k = 9k$</p><p>Therefore: $a = 5k$, $b = 6k$, $c = 7k$</p><p><strong>Step 3:</strong> Find inradius from inscribed circle area:</p><p>$\pi r^2 = \frac{8\pi}{3} \Rightarrow r^2 = \frac{8}{3} \Rightarrow r = \frac{2\sqrt{2}}{\sqrt{3}} = \frac{2\sqrt{6}}{3}$</p><p><strong>Step 4:</strong> Using Heron's formula with $s = 9k$:</p><p>$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{9k \cdot 4k \cdot 3k \cdot 2k} = \sqrt{216k^4} = 6\sqrt{6}k^2$</p><p><strong>Step 5:</strong> Use $\Delta = rs$:</p><p>$6\sqrt{6}k^2 = \frac{2\sqrt{6}}{3} \cdot 9k$</p><p>$6\sqrt{6}k^2 = 6\sqrt{6}k \Rightarrow k = 1$</p><p><strong>Step 6:</strong> Therefore: $a = 5$, $b = 6$, $c = 7$, $s = 9$, $\Delta = 6\sqrt{6}$, $r = \frac{2\sqrt{6}}{3}$</p><p><strong>Step 7:</strong> Verify options using cosine rule $\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{36+49-25}{84} = \frac{5}{7}$</p><p>Similarly: $\cos B = \frac{1}{7}$, $\cos C = -\frac{1}{14}$</p><p>∴ Answer: ACD</p>