The shortest distance from the plane $$12x+4y+3z=327$$ to the sphere

$${x^2} + {y^2} + {z^2} + 4x - 2y - 6z = 155$$ is

The two lines $$x=ay+b,z=cy+d$$ and $$x = a'y + b',z = c'y + d'$$ will be perpendicular, if and only if :

The lines $${{x - 2} \over 1} = {{y - 3} \over 1} = {{z - 4} \over { - k}}$$ and $${{x - 1} \over k} = {{y - 4} \over 2} = {{z - 5} \over 1}$$ are coplanar if :

The radius of the circle in which the sphere

$${x^2} + {y^2} + {z^2} + 2x - 2y - 4z - 19 = 0$$ is cut by the plane

$$x+2y+2z+7=0$$ is