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

$\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b}\times \overrightarrow{r}\Rightarrow \overrightarrow{r}\times \left(\overrightarrow{a}+\overrightarrow{b}\right)=0$

$\overrightarrow{r}=\overrightarrow{\lambda }\left(\overrightarrow{a}+\overrightarrow{b}\right)\Rightarrow \overrightarrow{r}=\overrightarrow{\lambda }\left(\hat{\mathrm{i}}+2\hat{\mathrm{j}}-3\hat{\mathrm{k}}+2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+5\hat{\mathrm{k}}\right)$

$\overrightarrow{r}=\overrightarrow{\lambda }\left(3\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}}\right) ...\left(1\right)$

$\overrightarrow{r}·(\alpha \hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}})=3$

Put $\overrightarrow{r}$ from $\left(1\right)$

$\alpha \lambda =1 ...\left(2\right)$

$\overrightarrow{r}·\left(2\hat{\mathrm{i}}+5\hat{\mathrm{j}}-\alpha \hat{\mathrm{k}}\right)=-1$

Put $\overrightarrow{r}$ from (1)

$2\lambda \alpha -\lambda =1 ...\left(3\right)$

Solve $\left(2\right)$ and $\left(3\right)$

$\alpha =1,\lambda =1$

$\Rightarrow \overrightarrow{r}=3\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}}$

$|\overrightarrow{r}{|}^2=14 \& \alpha =1$

$\alpha +|\overrightarrow{r}{|}^2=15$