Let the position vectors of points ' A ' and ' B ' be i ^ + j ^ + k ^ and 2 i ^ + j ^ + 3 k…

Position vector of $P$ is $\overrightarrow{OP}=\frac{\overrightarrow{a}+\lambda \overrightarrow{b}}{\lambda +1}$ $$$\because \overrightarrow{\mathrm{OB}}·\overrightarrow{\mathrm{OP}}-3|\overrightarrow{\mathrm{OA}}\times \overrightarrow{\mathrm{OP}}{|}^2=6$

$\Rightarrow \overrightarrow{b}·\left(\frac{\overrightarrow{a}+\lambda \overrightarrow{b}}{\lambda +1}\right)-3{\left|\overrightarrow{a}\times \left(\frac{\overrightarrow{a}+\lambda \overrightarrow{b}}{\lambda +1}\right)\right|}^2=6$

 $\Rightarrow \frac{\overrightarrow{a}·\overrightarrow{b}+\lambda |\overrightarrow{b}{|}^2}{\lambda +1}-\frac{3{\lambda }^2}{(\lambda +1{)}^2}|\overrightarrow{a}\times \overrightarrow{b}{|}^2=6$

$\Rightarrow \frac{6+\lambda .14}{\lambda +1}-\frac{3{\lambda }^2}{(\lambda +1{)}^2}·6=6$

$\Rightarrow \frac{18{\lambda }^2}{(\lambda +1{)}^2}+6=6+\frac{8\lambda }{\lambda +1}$

$\Rightarrow 18{\left(\frac{\lambda }{\lambda +1}\right)}^2-\frac{8\lambda }{\lambda +1}=0$ $\left(\frac{\lambda }{\lambda +1}\ne 0\right)$

$\Rightarrow 10\lambda =8\Rightarrow \lambda =0.8$