Let $\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{b}=2 \hat{i}-3 \hat{j}+5 \hat{k}$ If $\vec{r} \times \vec{a}=\vec{b} \times \vec{r} \cdot(a \hat{i}+2 \hat{j}+\hat{k})=3$ and $\vec{r} \cdot(2 \hat{i}+5 \hat{j}-\alpha \hat{k})=-1, \alpha \in R$, then the value of $\alpha+|\overrightarrow{\mathrm{r}}|^2$ is equal to :