Let the position vectors of points 'A' and 'B' be $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$, respectively. A point $' P^{\prime}$ divides the line segment $A B$ internally in the ratio $\lambda: 1(\lambda>0)$. If $\mathrm{O}$ is the origin and $\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OP}}-3|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OP}}|^{2}=6$, then $\lambda$ is equal to________.
Using section formula we get
$\overline{\mathrm{OP}}=\frac{2 \lambda+1}{\lambda+1} \hat{\mathrm{i}}+\frac{\lambda+1}{\lambda+1} \hat{\mathrm{j}}+\frac{3 \lambda+1}{\lambda+1} \hat{\mathrm{k}}$
Now
$\overline{\mathrm{OB}} \cdot \overline{\mathrm{OP}}=\frac{4 \lambda+2+\lambda+1+9 \lambda+3}{\lambda+1}$
$=\frac{14 \lambda+6}{\lambda+1}$
$\overline{\mathrm{OA}} \times \overline{\mathrm{OP}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 1 & 1 \\ \frac{2 \lambda+1}{\lambda+1} & 1 & \frac{3 \lambda+1}{\lambda+1}\end{array}\right|$
$=\frac{2 \lambda+1}{\lambda+1} \hat{\mathrm{i}}+\frac{-\lambda}{\lambda+1} \hat{\mathrm{j}}+\frac{-\lambda}{\lambda+1} \hat{\mathrm{k}}$
$|\overline{\mathrm{OA}} \times \overline{\mathrm{OP}}|^{2}=\frac{(2 \lambda+1)^{2}+\lambda^{2}+\lambda^{2}}{(\lambda+1)^{2}}$
$=\frac{6 \lambda^{2}+1}{(\lambda+1)^{2}}$
$\Rightarrow \frac{14 \lambda+6}{\lambda+1}-3 \times \frac{\left(6 \lambda^{2}+1\right)}{(\lambda+1)^{2}}=6$
$\Rightarrow 10 \lambda^{2}-8 \lambda=0$
$\Rightarrow \lambda=0, \frac{8}{10}=0.8$
$\Rightarrow \lambda=0.8$
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