Interesting !!!

We know that if gcd(a,b,c)=1
Then a = 2st , b = s²-t² and c = s²+t²

Hence area = a*b/2
=st(s+t)(s-t)

Now if either s or t ≡ 0(mod 2) then area is divisible by 2
else s+t ≡ 0(mod 2).

Hence area is divisible by 2.

Further if either s or t ≡ 0(mod 3) then area is divisible by 2 .

Otherwise if s ≡ 1(mod 3) & t ≡ -1(mod 3)
or if s ≡ -1(mod 3) & t ≡ 1(mod 3)
then s+t ≡ 0(mod 3)

And if s ≡ 1(mod 3) & t ≡ 1(mod 3)
or if s ≡ -1(mod 3) & t ≡ -1(mod 3)
then s-t ≡ 0(mod 3)

Hence area is Divisible by 6.

Now if gcd(a,b,c)=d
Then let a₁ =a/d
b₁ =b/d
and c₁ =c/d

gcd ( a₁ ,b₁ ,c₁)=1
Now a₁b ₁ c₁ forms a triangle similar to abc and has an area divisible by 6.Let it be 6*A.

Hence area of abc = d²*6*A whch is divisible by 6.