Let position vector of P is $\overline{\mathrm{OP}} \frac{\lambda \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}}}{\lambda+1}$ Give $\overline{\mathrm{OB}} \cdot \overline{\mathrm{OP}}-3|\overline{\mathrm{OA}} \cdot \overline{\mathrm{OP}}|^2=6$ $$ \begin{aligned} & \Rightarrow \quad \overline{\mathrm{b}} \cdot\left(\frac{\lambda \overline{\mathrm{~b}}+\overline{\mathrm{a}}}{\lambda+1}\right)-3\left|\overline{\mathrm{a}} \times \frac{\lambda \overline{\mathrm{b}}+\overline{\mathrm{a}}}{\lambda+1}\right|^2=6 \\ & \Rightarrow \quad \frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}+\lambda|\overline{\mathrm{b}}|^2}{\lambda+1} \frac{-3 \lambda^2}{(\lambda+1)}|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|^2=6 \\ & \quad(\because \overline{\mathrm{a}} \times \overline{\mathrm{b}}-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & \Rightarrow \quad \frac{6+14 \lambda}{\lambda+1}-\frac{18 \lambda^2}{(\lambda+1)^2}=6 \\ & \Rightarrow \quad 6+\frac{8 \lambda}{\lambda+1}-\frac{18 \lambda^2}{(\lambda+1)^2}=6 \end{aligned} $$ Let $\frac{\lambda}{\lambda+1}=t$ $$ \begin{array}{ll} & 18 t^2-8 t=0 \\ & t=0, \frac{4}{9} \\ \therefore & \frac{\lambda}{\lambda+1}=\frac{4}{9} \\ \therefore & \lambda=\frac{4}{5}=0.8 \end{array} $$