A line with direction ratios $〈2,1,2〉$ intersect each of the lines  $x=y+a=z$ and  $x+a=2y=2z$  the co-ordinates of each of the points of intersection are

The given lines are$\frac{x}{1}=\frac{y+a}{1}=\frac{z}{1},\frac{x+a}{2}=\frac{y}{1}=\frac{z}{1}$

Suppose that the line $L$ intersect the above two lines at $P,Q$

The direction rations of $PQ$ are $〈2,1,2〉$

The points   $P\left(s,s-a,s\right)$and  $Q\left(2t-a,t,t\right)$, these are general points on the given two lines.

 Direction ratios of  $PQ$are  $〈2t-a-s,t-s+a,t-s〉$, these are proportional to $〈2,1,2〉$

It implies that$\frac{2t-a-s}{2}=\frac{t-s+a}{1}=\frac{t-s}{2}$

Hence, $s=3a,t=a$

Therefore, the points of intersections are   $\boxed{P\left(3a,2a,3a\right),Q\left(a,a,a\right)}$