The quotient space of $\mathbb{S}^1\times [0,1]$ obtained by identifying points in the circle $\mathbb{S}^1\times \{0\}$ that differ by $2π /3$ rotation and identifying points in the circle $\mathbb{S}^1\times \{1\}$ that differ by $2π/2$ rotation.
My attempt: I will calculate the homology groups using the succession of mayer vietoris, for this, we call $X=\mathbb{S}^1\times [0,1]/{\sim}$, $U=X-(\mathbb{S}^1\times \{0\})$ and $V=X-(\mathbb{S}^1\times \{1\})$ then we will have that $U\simeq \mathbb{S}^1$, $V\simeq \mathbb{S}^1$ and $U\cap V$ is contractile, with these choices I get to that $H_0(X)\cong\mathbb{Z}$, $H_1(X)\cong\mathbb{Z}^2$, $H_n(X)=0$ if $n>1$, but I think I am wrong, could someone tell me if This is good? Thank you.
Consider the half-cylinders $\mathbb{S}^1\times [\frac{1}{2} ,1]/\sim $ and $\mathbb{S}^1\times [0,\frac{1}{2} ]/\sim $ and use the Mayer-Vietoris sequence, we will obtain $$ H_{\bullet } (\mathbb{S}^1\times\{\frac{1}{2}\} )\rightarrow H_{\bullet } (\mathbb{S}^1\times [\frac{1}{2} ,1]/\sim )\oplus H_{\bullet } (\mathbb{S}^1\times [0,\frac{1}{2} ]/\sim )\rightarrow H_{\bullet } (\mathbb{S}^1\times [0,1]/\sim )\xrightarrow{+1} $$ and observe that the first term is just $\mathbb{Z} $ when $\bullet =1$ and $0$ otherwise, with generator mapping onto $(3,2)$ via the given homomorphism, and hence the long exact sequence gives $H_{\bullet } (\mathbb{S}^1\times [0,1]/\sim )\cong (\mathbb{Z}\oplus\mathbb{Z} )/(3,2)\mathbb{Z}\cong\mathbb{Z}$.(Thanks for the comment, I made a mistake here, and now it's corrected.)