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If $\mathbf{P Q}+\mathbf{Q R}=\left(2 \lambda^2-5\right) \mathbf{R P}$ then, $\lambda$ is equal to
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$\pm \sqrt{2}$
$\begin{aligned} & \text { Given, } \mathbf{P Q}+\mathbf{Q R}=\left(2 \lambda^2-5\right) \mathbf{R} \mathbf{P} \\ & \Rightarrow \quad \mathbf{P R}=\left(2 \lambda^2-5\right) \mathbf{R P} \\ & \Rightarrow \quad-\mathbf{R P}=\left(2 \lambda^2-5\right) \mathbf{R P} \\ & \therefore \quad 2 \lambda^2-5=-1 \Rightarrow \lambda^2=2 \Rightarrow \lambda= \pm \sqrt{2} \\ & \end{aligned}$
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