Distance between two parallel planes

$$\,2x + y + 2z = 8$$ and $$4x + 2y + 4z + 5 = 0$$ is :

The intersection of the spheres
$${x^2} + {y^2} + {z^2} + 7x - 2y - z = 13$$ and
$${x^2} + {y^2} + {z^2} - 3x + 3y + 4z = 8$$
is the same as the intersection of one of the sphere and the plane

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 :