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 :

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 :