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.

Performing elementary row and column operations you can bring your matrix of vectors to the form $$\begin{pmatrix}2&0&0\\0&1&0\\0&0&1\end{pmatrix}$$ so the quotient is $\Bbb Z_2$. In the corrected answer by Michael Albanese, you would end up with a $0$ instead of a $2$ (carry out the calculation yourself), so the quotient is indeed $\Bbb Z$. Note that the first homology group is zero, not $\Bbb Z$, unless the 2-simplex is not supposed to be there.
