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 point of intersection are given by

A general point on the line $x=y+a=z$ is $(t_1,t_1-a,t_1)$.
A general point on the line $\frac{x+a}{2}=y=z$ is $(2t_2-a,t_2,t_2)$.
The direction ratios of the line segment joining these points are:
$(2t_2-a-t_1,t_2-(t_1-a),t_2-t_1)=(2t_2-a-t_1,t_2-t_1+a,t_2-t_1)$.
Given these are proportional to $2,1,2$:$2t_2-a-t_1=2k$ $t_2-t_1+a=k$ $t_2-t_1=2k$ From (2) and (3): $(t_2-t_1)+a=k\Rightarrow 2k+a=k\Rightarrow k=-a$.
Substituting $k=-a$ in (3): $t_2-t_1=-2a\Rightarrow t_1=t_2+2a$.
Substitute $t_1$ and $k$ in (1): $2t_2-a-(t_2+2a)=2(-a)$
$t_2-3a=-2a\Rightarrow t_2=a$.
Then $t_1=a+2a=3a$.
The points are $(3a,3a-a,3a)=(3a,2a,3a)$ and $(2(a)-a,a,a)=(a,a,a)$.