1. Let the required line pass through the two intersection points, one on each given line, and have direction ratios proportional to $2:1:2$.

2. Parametrize the first line:

   Given $x=y+a=z$. Let x=z=t, y=t-a. So a general point on the first line is $P(t)=(t,t-a,t).$

3. Parametrize the second line:

   Given $x+a=2y=2z$. Let $x+a=2y=2z=s.$ Then $x=s-a,\quad y=\frac{s}{2},\quad z=\frac{s}{2}.$ So a general point on the second line is $Q(s)=\left(s-a,\frac{s}{2},\frac{s}{2}\right).$

4. Since the line joining $P$ and $Q$ has direction ratios proportional to $(2,1,2)$, we must have $Q-P \propto (2,1,2).$

   Compute: [ Q-P=\left(s-a-t,\frac{s}{2}-(t-a),\frac{s}{2}-t\right). ]

   Hence, $\left(s-a-t,\frac{s}{2}-t+a,\frac{s}{2}-t\right)=\lambda(2,1,2).$

5. Equate components:

   $s-a-t=2\lambda \quad ...(1)$ $\frac{s}{2}-t+a=\lambda \quad ...(2)$ $\frac{s}{2}-t=2\lambda \quad ...(3)$

6. From (3), $2\lambda=\frac{s}{2}-t.$ Using this in (1): $s-a-t=\frac{s}{2}-t$ $s-a=\frac{s}{2}$ $\frac{s}{2}=a$ $s=2a.$

7. Substitute $s=2a$ into (3): $a-t=2\lambda.$ From (2): $a-t+a=\lambda$ $2a-t=\lambda.$

   Therefore, $a-t=2(2a-t)$ $a-t=4a-2t$ $t=3a.$

8. Now find the points.

   On the first line: $P=(t,t-a,t)=(3a,2a,3a).$

   On the second line: $Q=\left(s-a,\frac{s}{2},\frac{s}{2}\right)=\left(2a-a,a,a\right)=(a,a,a).$

9. So the two intersection points are $\boxed{(3a,2a,3a)\text{ and }(a,a,a)}.$

10. Comparing with the options, this is Option B.

11. Comparison with stored correct answer:

Stored correct answer = B.

Our derived answer also = B.

Hence, they agree.