If equation of the plane that contains the point $$(-2,3,5)$$ and is perpendicular to each of the planes $$2 x+4 y+5 z=8$$ and $$3 x-2 y+3 z=5$$ is $$\alpha x+\beta y+\gamma z+97=0$$ then $$\alpha+\beta+\gamma=$$

Let the image of the point $$\mathrm{P}(1,2,6)$$ in the plane passing through the points $$\mathrm{A}(1,2,0), \mathrm{B}(1,4,1)$$ and $$\mathrm{C}(0,5,1)$$ be $$\mathrm{Q}(\alpha, \beta, \gamma)$$. Then $$\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)$$ is equal to :

Let the line $$\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$$ intersect the lines $$\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$$ and $$\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$$ at the points $$\mathrm{A}$$ and $$\mathrm{B}$$ respectively. Then the distance of the mid-point of the line segment $$\mathrm{AB}$$ from the plane $$2 x-2 y+z=14$$ is :

The shortest distance between the lines $${{x + 2} \over 1} = {y \over { - 2}} = {{z - 5} \over 2}$$ and $${{x - 4} \over 1} = {{y - 1} \over 2} = {{z + 3} \over 0}$$ is :