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

If the straight lines
$$x=1+s,y=-3$$$$ - \lambda s,$$ $$z = 1 + \lambda s$$ and $$x = {t \over 2},y = 1 + t,z = 2 - t,$$ with parameters $$s$$ and $$t$$ respectively, are co-planar, then $$\lambda $$ equals :

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 :