Given lines are
$$
\begin{aligned}
& \qquad \frac{x-1}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda \text { (let) } \\
& \Rightarrow \quad x=\lambda+1, y=2 \lambda+1, z=3 \lambda+2 \\
& \text { and } \frac{x-5}{2}=\frac{y-2}{1}=z=\mu \text { (let) } \\
& \Rightarrow x=2 \mu+5, y=\mu+2, z=\mu
\end{aligned}
$$
For point of intersection, we have
$\lambda+1=2 \mu+5$ and $2 \lambda+1=\mu+2$ and $3 \lambda+2=\mu$
$\lambda-2 \mu=4$
and $2 \lambda-\mu=1$
and $3 \lambda-\mu=-2$
..(iii)
On solving Eqs. (ii) and (iii), we get
$$
\begin{aligned}
& \Rightarrow \quad \lambda=-3 \\
&
\end{aligned}
$$
Now, put $\lambda=-3$ and $\mu=-7$ in Eq. (i), we get
$$
-3+14=11 \neq 4
$$
Hence, the lines do not intersect.
Hence, option (4) is correct.