If the line $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over 4}$$ and $${{x - 3} \over 1} = {{y - k} \over 2} = {z \over 1}$$ intersect, then $$k$$ is equal to :

If the angle between the line $$x = {{y - 1} \over 2} = {{z - 3} \over \lambda }$$ and the plane

$$x+2y+3z=4$$ is $${\cos ^{ - 1}}\left( {\sqrt {{5 \over {14}}} } \right),$$ then $$\lambda $$ equals :

Statement - 1 : The point $$A(1,0,7)$$ is the mirror image of the point

$$B(1,6,3)$$ in the line : $${x \over 1} = {{y - 1} \over 2} = {{z - 2} \over 3}$$

Statement - 2 : The line $${x \over 1} = {{y - 1} \over 2} = {{z - 2} \over 3}$$ bisects the line

segment joining $$A(1,0,7)$$ and $$B(1, 6, 3)$$

Statement-1 : The point $$A(3, 1, 6)$$ is the mirror image of the point $$B(1, 3, 4)$$ in the plane $$x-y+z=5.$$

Statement-2 : The plane $$x-y+z=5$$ bisects the line segment joining $$A(3, 1, 6)$$ and $$B(1, 3, 4).$$