Showing $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}/(1,-1,0)\mathbb{Z}+(0,1,1)\mathbb{Z}+(1,0,-1)\mathbb{Z}\cong \mathbb{Z}$ I have to prove that if $V_K = \{v_0, v_1, v_2\}$ and $K = \{\{v_0\}, \{v_1\}, \{v_2\}, \{v_0, v_1\}, \{v_0, v_2\}, \{v_1, v_2\}\}$ then $H_q(K, \mathbb{Z})\cong  \mathbb{Z}$ for $q = 0, 1$. 
Already I proved that for $q=1$. For $H_0(K, \mathbb{Z})=\ker\delta_0/\operatorname{Im}\delta_1$, $\delta_0 : C_0(K)\to 0$, so it is obvious that $\ker\delta_0 = C_0(K)$. Also, 
$$\operatorname{Im}\delta_1 = \{(a + c)v_0 + (-a + b)v_1 + (b - c)v_2 \mid  a, b, c \in \mathbb{Z}\}$$
according to the definition of $\delta_1$.
Finally, this set can be written as follows:
$$\operatorname{Im}\delta_1 = (1, -1, 0)\mathbb{Z} + (0, 1, 1)\mathbb{Z} + (1, 0, -1)\mathbb{Z}$$ 
so the only thing I have to prove is that
$$\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}/(1,-1,0)\mathbb{Z}+(0,1,1)\mathbb{Z}+(1,0,-1)\mathbb{Z}\cong \mathbb{Z}.$$
All I can say is that the vectors $e_i$ don't belongs to this subset.

I believe you got a sign wrong in you expression for $\operatorname{Im}\delta_1$. You should have
$$\operatorname{Im}\delta_1 = \{(a + c)v_0 + (-a + b)v_1 + (\color{red}{-b} - c)v_2 \mid a, b, c \in \mathbb{Z}\}$$
as $\delta_1\{v_0, v_2\} = 1\{v_0\} \color{red}{- 1}\{v_2\}$, so $\operatorname{Im}\delta = (1, -1, 0)\mathbb{Z} + (1, 0, -1)\mathbb{Z} + (0, 1, -1)\mathbb{Z}$. 
Now note that the group homomorphism $\varphi : \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z} \to \mathbb{Z}$ given by $\varphi(x, y, z) = x + y + z$ is surjective and $\operatorname{Im}\delta_1 \subseteq \ker\varphi$. If $(x, y, z) \in \ker\varphi$, then $x + y + z = 0$ so $z = -x - y$. As 
$$(x, y, z) = (x, y, -x - y) = x(1, 0, -1) + y(0, 1, -1) \in \operatorname{Im}\delta_1,$$ 
we see that $\ker\varphi \subseteq\operatorname{Im}\delta_1$, and therefore $\ker\varphi = \operatorname{Im}\delta_1$.
So by the first isomorphism theorem,
$$\frac{\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}}{(1, -1, 0)\mathbb{Z} + (1, 0, -1)\mathbb{Z} + (0, 1, -1)\mathbb{Z}} \cong \mathbb{Z}.$$ 
