Let $$\alpha, \beta, \gamma, \delta \in \mathbb{Z}$$ and let $$A(\alpha, \beta), B(1, 0), C(\gamma, \delta)$$ and $$D(1, 2)$$ be the vertices of a parallelogram $$ABCD$$. If $$AB = \sqrt{10}$$ and the points $$A$$ and $$C$$ lie on the line $$3y = 2x + 1$$, then $$2(\alpha + \beta + \gamma + \delta)$$ is equal to

ABCD is a parallelogram with B(1,0), D(1,2). Midpoint of BD = (1,1) = midpoint of AC.

A(α,β) and C(γ,δ) on line 3y=2x+1. Since midpoint of AC = (1,1): α+γ=2, β+δ=2.

AB=√10: (α-1)²+β²=10. A on 3β=2α+1: β=(2α+1)/3.

(α-1)²+(2α+1)²/9=10. 9(α-1)²+(2α+1)²=90. 9α²-18α+9+4α²+4α+1=90. 13α²-14α-80=0.

α=(14±√(196+4160))/26=(14±66)/26. α=80/26=40/13 or α=-2.

If α=-2: β=(-4+1)/3=-1. γ=4, δ=3. Check 3(3)=9, 2(4)+1=9 ✓.

If α=40/13: β=(80/13+1)/3=93/(13·3)=31/13. γ=2-40/13=-14/13. δ=2-31/13=-5/13.

Both solutions work. With integer α,β,γ,δ: α=-2,β=-1,γ=4,δ=3.

2(α+β+γ+δ)=2(-2-1+4+3)=2(4)=8.

The answer is Option (4): $$\boxed{8}$$.