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.

The quotient is not equal to $\Bbb Z$: it is equal to $\Bbb Z/2$.  In particular, to correct the comment by p Groups:
$f(1, -1, 0)=0$ implies $a-b=0$, so $a=b$.  Similarly, $a-c=0$ implies $a=c$.  Finally, $b+c=0$ implies $2a=0$.  So the quotient is just $a\Bbb Z/2a\Bbb Z$, for example with representatives $\{(0, 0, 0), (1, 0, 0)\}$.
In particular, $(0, 0, 1) = (1, 0, 0) - (1, 0, -1)$, and $(0, 1, 0) = (1, 0, 0) - (1, -1, 0)$.
