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

Let the point, on the line passing through the points $$P(1,-2,3)$$ and $$Q(5,-4,7)$$, farther from the origin and at a distance of 9 units from the point $$P$$, be $$(\alpha, \beta, \gamma)$$. Then $$\alpha^2+\beta^2+\gamma^2$$ is equal to :