Let $$A(-2, -1)$$, $$B(1, 0)$$, $$C(\alpha, \beta)$$ and $$D(\gamma, \delta)$$ be the vertices of a parallelogram $$ABCD$$. If the point $$C$$ lies on $$2x - y = 5$$ and the point $$D$$ lies on $$3x - 2y = 6$$, then the value of $$|\alpha + \beta + \gamma + \delta|$$ is equal to

ABCD parallelogram. A(-2,-1),B(1,0). Mid of AC=mid of BD. C(α,β) on 2x-y=5. D(γ,δ) on 3x-2y=6.

D=A+C-B=(α-3,β-1). D on 3x-2y=6: 3(α-3)-2(β-1)=6 → 3α-9-2β+2=6 → 3α-2β=13.

C on 2x-y=5: 2α-β=5 → β=2α-5.

3α-2(2α-5)=13 → 3α-4α+10=13 → -α=3 → α=-3. β=-11.

γ=α-3=-6, δ=β-1=-12.

|α+β+γ+δ|=|-3-11-6-12|=|-32|=32.

The answer is $$\boxed{32}$$.