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 $$\mathrm{PQR}$$ is $$\lambda$$ and $$\lambda^2=14 \mathrm{~K}$$, then $$\mathrm{K}$$ is equal to :

If $$A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1)$$ and $$D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$$ are the vertices of a quadrilateral $$A B C D$$, then its area is

The shortest distance between the lines $$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$$ and $$\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$$ is

Let $$(\alpha, \beta, \gamma)$$ be the image of the point $$(8,5,7)$$ in the line $$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}$$. Then $$\alpha+\beta+\gamma$$ is equal to :