Let $P=(1,2,a)$. The point $A$ on $L_1$ at parameter $\lambda$: $A=(3\lambda +1,4\lambda +2,b\lambda +a)$, and $B$ on $L_2$ at parameter $\mu$: $B=(\mu +1,4\mu +2,c\mu +a)$.

Both $A$ and $B$ lie on the central line $L:\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ (these are the foot points where the side lines hit $L$).

For $A$ on $L$: $\frac{3\lambda }{1}=\frac{4\lambda +2}{2}=\frac{b\lambda +a-1}{1}$. From the first equality: $3\lambda =2\lambda +1\Rightarrow \lambda =1$, then $a+b-1=3\Rightarrow a+b=4$. ...(I)

For $B$ on $L$: $\mu =\frac{4\mu +2}{2}=\frac{c\mu +a-1}{1}$. From the first equality: $\mu =2\mu +1\Rightarrow \mu =-1$, then $a-c-1=-1\Rightarrow a=c$. ...(II) So $B=(0,-2,0)$.

$|PA|=|PB|$ with $A=(4,6,4)$ (using $\lambda =1$ and (I)) and $B=(0,-2,0)$:

$|PA{|}^2=9+16+(a-4{)}^2=25+(a-4{)}^2$,

$|PB{|}^2=1+16+a^2=17+a^2$.

Equating: $25+a^2-8a+16=17+a^2\Rightarrow -8a+24=0\Rightarrow a=3$. Then $b=1$, $c=3$.

$a+b+c=3+1+3=7$. Option (1).