For the image of $P$ in line $L_1$ to land at $Q$ and the image of $Q$ in line $L_2$ to return to $P$, the foot of perpendicular from $P$ to $L_1$ must equal the midpoint of $PQ$; similarly for $L_2$.

Step 1 (foot on $L_1$): A general point on $L_1$ is $(2\mu ,\mu -a,\mu )$. The vector from this point to $P(a,2,a)$ is $(a-2\mu ,2-\mu +a,a-\mu )$ and must be perpendicular to direction $(2,1,1)$:

$2(a-2\mu )+(2-\mu +a)+(a-\mu )=0\Rightarrow 4a+2-6\mu =0\Rightarrow \mu =\frac{4a+2}{6}$.

For $\mu =a$ (so the foot is $(2a,0,a)$): $a=\frac{4a+2}{6}\Rightarrow 6a=4a+2\Rightarrow a=1$.

Then $Q=2(2a,0,a)-P=(4a-a,-2,2a-a)=(3a,-2,a)=(3,-2,1)$ (using $a=1$).

Step 2 (foot on $L_2$): For the image of $Q$ in $L_2$ to be $P$, the midpoint of $PQ$ must lie on $L_2$ and $PQ$ must be perpendicular to direction $(2,1,-5)$. Midpoint of $PQ$ is $(2,0,1)$. Substituting into $L_2$ with parameter:

$\frac{2-2b}{2}=\frac{0-1}{1}=\frac{1+2b}{-5}$.

From the middle equation, the parameter is $-1$. Then $\frac{2-2b}{2}=-1\Rightarrow b=2$.

Check the third: $\frac{1+4}{-5}=-1$ ✓.

$\therefore a+b=1+2=3$.