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

$\overrightarrow{r}\times \overrightarrow{a}-\overrightarrow{b}\times \overrightarrow{r}=\overrightarrow{0}\Rightarrow \overrightarrow{r}\times (\overrightarrow{a}+\overrightarrow{b})=\overrightarrow{0}\Rightarrow \overrightarrow{r}=\lambda (\overrightarrow{a}+\overrightarrow{b})=\lambda (3i-j+2k)$

$\Rightarrow \lambda (3i-j+2k).(\alpha i+2j+k)=3\Rightarrow 3\lambda \alpha -2\lambda +2\lambda =3\Rightarrow \lambda \alpha =1$

Also $\lambda (3i-j+2k).(2i+5j-\alpha k)=-1 \lambda (6-5-2\alpha )=-1,\lambda =1\alpha =1$