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

Let the point of intersection on the first line be

$$P=(t,t-a,t)$$

since

$$x=y+a=z$$

implies

$$x=z=t,\qquad y=t-a$$

Let the point of intersection on the second line be

$$Q=(2s-a,s,s)$$

since

$$x+a=2y=2z=2s$$

The given line joining $$P$$ and $$Q$$ has direction ratios proportional to

$$2,1,2$$

Therefore,

$$\overrightarrow{PQ}=(2s-a-t,\ s-t+a,\ s-t)$$

must satisfy

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

From

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

we get

$$2(s-t+a)=s-t$$

$$s-t+2a=0$$

$$s-t=-2a$$

From

$$\frac{2s-a-t}{2}=\frac{s-t}{2}$$

we get

$$2s-a-t=s-t$$

$$s=a$$

Hence,

$$t=3a$$

Substituting in the coordinates of $$P$$,

$$P=(3a,2a,3a)$$

Substituting in the coordinates of $$Q$$,

$$Q=(a,a,a)$$

Thus the two points of intersection are

$$\boxed{(3a,2a,3a)\ \text{and}\ (a,a,a)}$$