Let $\alpha , \quad \beta , \quad \gamma , \quad \delta \in Z$ 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 $3 y = 2 x + 1$, then $2 \alpha + \beta + \gamma + \delta$ is equal to\\
(1) 10\\
(2) 5\\
(3) 12\\
(4) 8