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 :

If the line $$\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z$$ makes a right angle with the line $$\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7}$$, then $$4 \lambda+9 \mu$$ is equal to :

Let $$\mathrm{d}$$ be the distance of the point of intersection of the lines $$\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$$ and $$\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$$ from the point $$(7,8,9)$$. Then $$\mathrm{d}^2+6$$ is equal to :

Let $$\mathrm{P}$$ be the point of intersection of the lines $$\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$$ and $$\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$$. Then, the shortest distance of $$\mathrm{P}$$ from the line $$4 x=2 y=z$$ is