JEE Math Practice Question
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 $$\mathrm{ABCD}$$. If $$A B=\sqrt{10}$$ and the points $$\mathrm{A}$$ and $$\mathrm{C}$$ lie on the line $$3 y=2 x+1$$, then $$2(\alpha+\beta+\gamma+\delta)$$ is equal to
- A.8
- B.5
- C.12
- D.10
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