The line $$l_1$$ passes through the point (2, 6, 2) and is perpendicular to the plane $$2x+y-2z=10$$. Then the shortest distance between the line $$l_1$$ and the line $$\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$$ is :

Let a unit vector $$\widehat{O P}$$ make angles $$\alpha, \beta, \gamma$$ with the positive directions of the co-ordinate axes $$\mathrm{OX}$$, $$\mathrm{OY}, \mathrm{OZ}$$ respectively, where $$\beta \in\left(0, \frac{\pi}{2}\right)$$. If $$\widehat{\mathrm{OP}}$$ is perpendicular to the plane through points $$(1,2,3),(2,3,4)$$ and $$(1,5,7)$$, then which one of the following is true ?

The plane $$2x-y+z=4$$ intersects the line segment joining the points A ($$a,-2,4)$$ and B ($$2,b,-3)$$ at the point C in the ratio 2 : 1 and the distance of the point C from the origin is $$\sqrt5$$. If $$ab < 0$$ and P is the point $$(a-b,b,2b-a)$$ then CP$$^2$$ is equal to

If the lines $${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1}$$ and $${{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1}$$ intersect at the point P, then the distance of the point P from the plane $$z = a$$ is :