If $\lambda_{1} < \lambda_{2}$ are two values of $\lambda$ such that the angle between the planes $P_{1}: \vec{r}(3 \hat{i}-5 \hat{j}+\hat{k})=7$ and $P_{2}: \vec{r} \cdot(\lambda \hat{i}+\hat{j}-3 \hat{k})=9$ is $\sin ^{-1}\left(\frac{2 \sqrt{6}}{5}\right)$, then the square of the length of perpendicular from the point $\left(38 \lambda_{1}, 10 \lambda_{2}, 2\right)$ to the plane $P_{1}$ is ______________.

<p>${P_1}:\overrightarrow r \,.\,(3\widehat i - 5\widehat j + \widehat k) = 7$</p> <p>${P_2}:\overrightarrow r \,.\,(\lambda \widehat i + \widehat j - 3\widehat k) = 9$</p> <p>Let angle between P₁ and P₂ is $\theta$</p> <p>Then $$\cos \theta = {{3\lambda - 5 - 3} \over {\sqrt {35} \sqrt {{\lambda ^2} + 10} }}$$</p> <p>But $\sin \theta = {{2\sqrt 6 } \over 5}$</p> <p>$\therefore$ ${{{{(3\lambda - 8)}^2}} \over {35({\lambda ^2} + 10)}} = 1 - {{24} \over {25}}$</p> <p>$\Rightarrow 5(9{\lambda ^2} + 64 - 48\lambda ) = 7{\lambda ^2} + 70$</p> <p>$\Rightarrow 38{\lambda ^2} - 240\lambda + 250 = 0$</p> <p>$\Rightarrow 19{\lambda ^2} - 120\lambda + 125 = 0$</p> <p>$\Rightarrow (19\lambda - 25)(\lambda - 5) = 0$</p> <p>$\therefore$ ${\lambda _1} = {{25} \over {19}},{\lambda _2} = 5$</p> <p>So, point (50, 50, 2)</p> <p>$\therefore$ $d = {{|150 - 250 + 2 - 7|} \over {\sqrt {35} }} = 315$</p>