Let the position vectors of points ‘A’ and ‘B’ be  $\widehat{i}+\widehat{j}+\widehat{k}$  and $2\widehat{i}+\widehat{j}+2\widehat{k}$ , respectively. A point ‘P’ divides the line segment AB internally in the ratio  $\lambda :1(\lambda >0)$ . If O is the origin and $\overline{OB}.\overline{OP}-5{|\overline{OA}\times \overline{OP}|}^2=5$ , then 5  $\overline{OP}$  is equal to ______

Let  $m:n=\lambda :1$
$\overline{OP}=\frac{\lambda \overline{b}+\overline{a}}{\lambda +1},$   where  $\overline{OA}=\overline{a},\overline{OB}=\overline{b}$
So    $\overline{OB}.\overline{OP}-5{|\overline{OA}\times \overline{OP}|}^2=5$                           $\Rightarrow \overline{b}.\left(\frac{\lambda \overline{b}+\overline{a}}{\lambda +1}\right)-5{|\overline{a}\times \left(\frac{\lambda \overline{b}+\overline{a}}{\lambda +1}\right)|}^2=5$
 $$\begin{gathered} \Rightarrow \frac{\overline{a}.\overline{b}+\lambda {\left|\overline{b}\right|}^2}{\lambda +1}-\frac{5{\lambda }^2}{{(\lambda +1)}^2}{|\overline{a}\times \overline{b}|}^2=5 \\ \begin{array}{l}\Rightarrow \frac{5+9\lambda }{\lambda +1}-\frac{5{\lambda }^2}{{(\lambda +1)}^2}\times 2=5\\ \Rightarrow \frac{5+9\lambda }{\lambda +1}-5=\frac{10{\lambda }^2}{{(\lambda +1)}^2}\\ \Rightarrow \frac{4\lambda }{\lambda +1}=\frac{10{\lambda }^2}{{(\lambda +1)}^2}\Rightarrow 2(\lambda +1)=5\lambda \Rightarrow \lambda =\frac{2}{3}\end{array} \end{gathered}$$