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 :

Let two vertices of a triangle ABC be (2, 4, 6) and (0, $$-$$2, $$-$$5), and its centroid be (2, 1, $$-$$1). If the image of the third vertex in the plane $$x+2y+4z=11$$ is $$(\alpha,\beta,\gamma)$$, then $$\alpha\beta+\beta\gamma+\gamma\alpha$$ is equal to :

Let P be the point of intersection of the line $${{x + 3} \over 3} = {{y + 2} \over 1} = {{1 - z} \over 2}$$ and the plane $$x+y+z=2$$. If the distance of the point P from the plane $$3x - 4y + 12z = 32$$ is q, then q and 2q are the roots of the equation :