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Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the image of the point $\mathrm{Q}(3,-3,1)$ in the line $\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}$ and $\mathrm{R}$ be the point $(2,5,-1)$. If the area of the triangle $P Q R$ is $\lambda$ and $\lambda^2=14 K$, then $K$ is equal to :
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81
$\begin{aligned} & \mathrm{RQ}=\sqrt{1+64+4}=\sqrt{69} \\ & \overrightarrow{\mathrm{RQ}}=\hat{\ell}-8 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{RS}}=\hat{\ell}+\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ & \cos \theta=\frac{\overrightarrow{\mathrm{RQ}} \cdot \overrightarrow{\mathrm{RS}}}{|\overrightarrow{\mathrm{RQ}}||\overrightarrow{\mathrm{RS}}|}=\left|\frac{1-8-2}{\sqrt{69} \sqrt{3}}\right|=\frac{9}{3 \sqrt{23}} \\ & \cos \theta=\frac{3}{\sqrt{23}}=\frac{\mathrm{RS}}{\mathrm{RQ}}=\frac{\mathrm{RS}}{\sqrt{69}} \\ & \mathrm{RS}=3 \sqrt{3} \\ & \sin \theta=\frac{\sqrt{14}}{\sqrt{23}}=\frac{\mathrm{QS}}{\sqrt{69}}\end{aligned}$
$\begin{aligned} & \mathrm{QS}=\sqrt{42} \\ & \text { area }=\frac{1}{2} \cdot 2 \mathrm{QS} \cdot \mathrm{RS}=\sqrt{42} \cdot 3 \sqrt{3} \\ & \lambda=9 \sqrt{14} \\ & \lambda^2=81.14=14 \mathrm{k} \\ & \mathrm{k}=81\end{aligned}$
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