1. Use the parallelogram property

In parallelogram $ABCD$ with vertices in order $A,B,C,D$, the diagonals bisect each other. So midpoint of $AC$ equals midpoint of $BD$.

Given:

• $B(1,0)$
• $D(1,2)$

Midpoint of $BD$ is $\left(\frac{1+1}{2},\frac{0+2}{2}\right)=(1,1).$

Hence midpoint of $AC$ is also $(1,1)$, so $\frac{\alpha+\gamma}{2}=1, \qquad \frac{\beta+\delta}{2}=1.$ Therefore, $\alpha+\gamma=2, \qquad \beta+\delta=2.$ Thus, $\alpha+\beta+\gamma+\delta=4.$ So, $2(\alpha+\beta+\gamma+\delta)=8.$

This already suggests option A.

2. Check consistency with the other conditions

We are also given that $A$ and $C$ lie on the line $3y=2x+1.$

Since $A(\alpha,\beta)$ lies on it, $3\beta=2\alpha+1.$ Also, $AB=\sqrt{10}$ where $B(1,0)$, so $(\alpha-1)^2+\beta^2=10.$

Now test integer solutions.

From $3\beta=2\alpha+1 \implies \alpha=\frac{3\beta-1}{2},$ so $\beta$ must be odd.

Try odd integer values of $\beta$ such that distance condition may hold:

• If $\beta=1$, then $\alpha=1$ and $AB^2=(1-1)^2+1^2=1 \neq 10.$
• If $\beta=-1$, then $\alpha=-2$ and $AB^2=(-2-1)^2+(-1)^2=9+1=10.$ Works.
• If $\beta=3$, then $\alpha=4$ and $AB^2=(4-1)^2+3^2=9+9=18 \neq 10.$
• If $\beta=-3$, then $\alpha=-5$ and distance is too large.

Hence $A=(-2,-1).$

Using midpoint $(1,1)$ of $AC$, $\gamma=2-\alpha=2-(-2)=4,$ $\delta=2-\beta=2-(-1)=3.$ So $C=(4,3).$

Check: $C$ lies on $3y=2x+1$: $3(3)=9, \qquad 2(4)+1=9,$ correct.

Now compute: $\alpha+\beta+\gamma+\delta=-2-1+4+3=4,$ therefore $2(\alpha+\beta+\gamma+\delta)=2\cdot 4=8.$

3. Evaluate options

• A: $8$ ✅
• B: $5$ ❌
• C: $12$ ❌
• D: $10$ ❌

So the correct answer is A.