1. Write the lines in parametric form

For $L_1$: $\frac{x-1}{2}=\frac{y-3}{1}=\frac{z-2}{2}=t$ So a general point on $L_1$ is $P(1+2t,\,3+t,\,2+2t).$

For $L_2$: $\frac{x-2}{1}=\frac{y-2}{2}=\frac{z-3}{3}=s$ So a general point on $L_2$ is $Q(2+s,\,2+2s,\,3+3s).$

2. Use the direction ratios of $L_3$

Given $L_3$ has direction ratios $(1,-1,-2)$ and intersects $L_1$ at $P$ and $L_2$ at $Q$.

Hence the vector $\overrightarrow{PQ}$ must be parallel to $(1,-1,-2)$.

So, $\overrightarrow{PQ}=Q-P$ must satisfy $Q-P=\lambda(1,-1,-2)$ for some scalar $\lambda$.

Now, $Q-P=((2+s)-(1+2t),\,(2+2s)-(3+t),\,(3+3s)-(2+2t))$ $=(1+s-2t,\,-1+2s-t,\,1+3s-2t).$

Thus, $1+s-2t=\lambda \quad ...(1)$ $-1+2s-t=-\lambda \quad ...(2)$ $1+3s-2t=-2\lambda \quad ...(3)$

3. Solve for $s,t$

From (1) and (2): $1+s-2t=-( -1+2s-t)=1-2s+t$ $s-2t=-2s+t$ $3s=3t$ $s=t.$

Now use (1) and (3): $1+s-2t=-\frac{1}{2}(1+3s-2t).$ Since $s=t$, equation (1) gives $\lambda=1-t.$ Also from (3), $1+3t-2t=1+t=-2\lambda=-2(1-t).$ So, $1+t=-2+2t$ $t=3.$ Hence, $s=3.$

4. Find the points $P$ and $Q$

At $t=3$, $P=(1+2\cdot 3,\,3+3,\,2+2\cdot 3)=(7,6,8).$

At $s=3$, $Q=(2+3,\,2+2\cdot 3,\,3+3\cdot 3)=(5,8,12).$

5. Compute the length $PQ$

$\overrightarrow{PQ}=Q-P=(5-7,\,8-6,\,12-8)=(-2,2,4).$

Therefore, $PQ=\sqrt{(-2)^2+2^2+4^2}$ $=\sqrt{4+4+16}$ $=\sqrt{24}=2\sqrt{6}.$

6. Check with options

The length of the segment $PQ$ is $\boxed{2\sqrt{6}}.$ So the correct option is B.

7. Comparison with stored answer

Stored correct answer: B

Derived answer: B

They agree.