$\begin{array}{l}\text{ Let }\overrightarrow{a}=\hat{i}+2\hat{j}-3\hat{k}\text{ and }\overrightarrow{b}=2\hat{j}-3\hat{j}+5\hat{k}\text{ . If }\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b}\times \overrightarrow{r},\overrightarrow{r}·(\alpha \hat{i}+2\hat{j}+\hat{k})=3\text{ and }\\ \overrightarrow{r}.(2\hat{i}+5\hat{j}-\alpha \hat{k})=-1,\alpha \in R,\text{ Then the value of }\alpha +|\overrightarrow{r}{|}^2\text{ is equal to : }\end{array}$

$\begin{array}{l}\text{Given }\overline{r}\times \overline{a}=\overline{b}\times \overline{r}\\ \text{it implies }\overline{r}\times \left(\overline{a}+\overline{b}\right)=\overline{0}\\ \text{hence, }\overline{r}=k\left(\overline{a}+\overline{b}\right)\\ \\ =k\left(i+2j-3k+2i-3j+5k\right)\\ =k\left(3i-j+2k\right)\\ \\ \text{Given }\overline{r}\cdot \left(\alpha i+2j+k\right)=3\\ \Rightarrow k\left(3\alpha -2+2\right)=3\\ \Rightarrow k\left(\alpha \right)=1\mathrm{......}\left(1\right)\\ \\ \text{Given }\overline{r}\cdot \left(2i+5j-\alpha k\right)=-1\\ \Rightarrow k\left(6-5-2\alpha \right)=-1\\ \Rightarrow k\left(1-2\alpha \right)=-1.\mathrm{......}\left(2\right)\\ \\ \\ \text{From equations }\left(1\right)\text{ and }\left(2\right),\\ \alpha =1,k=1\\ \left|r\right|=\sqrt{9+1+4}=\sqrt{14}\\ \alpha +{\left|\overrightarrow{r}\right|}^2=1+14=\boxed{15}\end{array}$