Let A denote the event that a 6-digit integer formed by

Total cases :

$\underline{6} \cdot \underline{6} \cdot \underline{5} \cdot \underline{4} \cdot \underline{3} \cdot \underline{2}$

$n(s)=6 \cdot 6 !$

Favourable cases :

Number divisible by $3 \equiv$

Sum of digits must be divisible by 3

Case-I

$1,2,3,4,5,6$

Number of ways $=6 !$

Case-II

$0,1,2,4,5,6$

Number of ways $=5.5 !$

Case-III

$0,1,2,3,4,5$

Number of ways $=5.5 !$

$\mathrm{n}($ favourable $)=6 !+2 \cdot 5 \cdot 5 !$

$P=\frac{6 !+2 \cdot 5 \cdot 5 !}{6 \cdot 6 !}=\frac{4}{9}$