A line with direction cosines proportional to $$2,\,1,\,2$$   meets each of the lines $$x=y+a=z$$    and $$x+a=2y=2z.$$    The co-ordinates of each of the points of intersection are given by :

Solution :
Let a point on the line $$x=y+a=z$$    is $$\left( {\lambda ,\,\lambda - a,\,\lambda } \right)$$   and a point on the line $$x+a=2y=2z$$    is $$\left( {\mu - a,\,\frac{\mu }{2},\,\frac{\mu }{2}} \right),$$    then direction ratio of the line joining these points are $$\lambda - \mu + a,\,\lambda - a - \frac{\mu }{2},\,\lambda - \frac{\mu }{2}$$
If it respresents the required line, then
$$\frac{{\lambda - \mu + a}}{2} = \frac{{\lambda - a - \frac{\mu }{2}}}{1} = \frac{{\lambda - \frac{\mu }{2}}}{2}$$
on solving we get $$\lambda = 3a,\,\,\mu = 2a$$
$$\therefore $$ The required points of intersection are
$$\left( {3a,\,3a - a,\,3a} \right)$$    and $$\left( {2a - a,\,\frac{{2a}}{2},\,\frac{{2a}}{2}} \right)$$
or $$\left( {3a,\,2a,\,3a} \right)$$    and $$\left( {a,\,a,\,a} \right)$$