Let $$A$$ denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then probability of event $$A$$ is equal to:

We need to form a 6-digit integer using digits from $$\{0, 1, 2, 3, 4, 5, 6\}$$ without repetition, and the number must be divisible by 3.

The total sum of all 7 digits is $$0+1+2+3+4+5+6 = 21$$. To form a 6-digit number, we exclude one digit. For the number to be divisible by 3, the sum of the 6 chosen digits must be divisible by 3. Since $$21$$ is divisible by 3, the excluded digit must also be divisible by 3. The digits divisible by 3 from the set are $$\{0, 3, 6\}$$.

The total number of valid 6-digit integers (no repetition, first digit non-zero) from 7 digits: we exclude one digit out of 7. If 0 is excluded, the remaining digits are $$\{1,2,3,4,5,6\}$$, giving $$6! = 720$$ six-digit numbers. If a non-zero digit is excluded, the remaining 6 digits include 0, giving $$6! - 5! = 720 - 120 = 600$$ valid six-digit numbers (subtracting those starting with 0). Total = $$720 + 6 \times 600 = 720 + 3600 = 4320$$.

Now we count favorable cases where we exclude a digit divisible by 3. Case 1: Exclude 0. Remaining digits $$\{1,2,3,4,5,6\}$$, sum = 21, divisible by 3. Number of 6-digit numbers = $$6! = 720$$. Case 2: Exclude 3. Remaining $$\{0,1,2,4,5,6\}$$, sum = 18, divisible by 3. Number of valid 6-digit numbers = $$6! - 5! = 600$$. Case 3: Exclude 6. Remaining $$\{0,1,2,3,4,5\}$$, sum = 15, divisible by 3. Number of valid 6-digit numbers = $$6! - 5! = 600$$.

Total favorable = $$720 + 600 + 600 = 1920$$.

Probability = $$\frac{1920}{4320} = \frac{4}{9}$$.

The answer is $$\frac{4}{9}$$, which corresponds to Option (2).