Question:

<p><strong>Step 1: Setup and Residue Classes</strong><br/>Consider 3n consecutive integers. When divided by 3, they give remainders 0, 1, 2. Since we have 3n consecutive integers, exactly n integers fall into each residue class modulo 3.</p><p><strong>Step 2: Total Ways to Select 3 Integers</strong><br/>Total number of ways to select 3 integers from 3n integers: $$\binom{3n}{3} = \frac{3n(3n-1)(3n-2)}{6}$$</p><p><strong>Step 3: Favorable Outcomes (Sum ≡ 0 mod 3)</strong><br/>For three integers to have sum divisible by 3, we need: <ul><li><strong>Case 1:</strong> All three from the same residue class (all ≡ 0, all ≡ 1, or all ≡ 2 mod 3)</li> <li><strong>Case 2:</strong> One from each residue class (one ≡ 0, one ≡ 1, one ≡ 2 mod 3)</li> </ul></p><p><strong>Step 4: Count Case 1 (All from same class)</strong><br/>Number of ways to choose 3 from the same residue class: $$3 \times \binom{n}{3} = 3 \times \frac{n(n-1)(n-2)}{6} = \frac{n(n-1)(n-2)}{2}$$</p><p><strong>Step 5: Count Case 2 (One from each class)</strong><br/>Number of ways to choose one from each of the three residue classes: $$\binom{n}{1} \times \binom{n}{1} \times \binom{n}{1} = n^3$$</p><p><strong>Step 6: Total Favorable Outcomes</strong><br/>$$\text{Favorable} = \frac{n(n-1)(n-2)}{2} + n^3 = \frac{n(n-1)(n-2) + 2n^3}{2}$$<br/>$$= \frac{n[(n-1)(n-2) + 2n^2]}{2} = \frac{n[n^2 - 3n + 2 + 2n^2]}{2}$$<br/>$$= \frac{n[3n^2 - 3n + 2]}{2}$$</p><p><strong>Step 7: Calculate Probability</strong><br/>$$P = \frac{\text{Favorable}}{\text{Total}} = \frac{\frac{n(3n^2-3n+2)}{2}}{\frac{3n(3n-1)(3n-2)}{6}}$$<br/>$$= \frac{n(3n^2-3n+2)}{2} \times \frac{6}{3n(3n-1)(3n-2)}$$<br/>$$= \frac{3n(3n^2-3n+2)}{3n(3n-1)(3n-2)} = \frac{3n^2-3n+2}{(3n-1)(3n-2)}$$</p><p><strong>∴ Answer: A</strong></p>