Check whether the lines $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}\text{ and }\frac{\mathrm{x}-4}{5}=\frac{\mathrm{y}-1}{2}=\mathrm{z}$ are skew or not.

Given: $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}\dots \dots \left(1\right)$ and $\frac{\mathrm{x}-4}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-0}{1}........\left(2\right)$
So, ${\mathrm{x}}_1=1,{\mathrm{y}}_1=2,{\mathrm{z}}_1=3$ and ${\mathrm{x}}_2=4,{\mathrm{y}}_2=1,{\mathrm{z}}_2=0$ also ${\mathrm{l}}_1=2,{\mathrm{m}}_1=3,{\mathrm{n}}_1=4$ and ${\mathrm{l}}_2=5,{\mathrm{m}}_2=2,{\mathrm{n}}_2=1$
Now for to check the lines are skew then it should not be coplanar,
So, for coplanarity,
$\left|\begin{array}{ccc}{\mathrm{x}}_2-{\mathrm{x}}_1& {\mathrm{y}}_2-{\mathrm{y}}_1& {\mathrm{z}}_2-{\mathrm{z}}_1\\ {\mathrm{l}}_1& {\mathrm{m}}_1& {\mathrm{n}}_1\\ {\mathrm{l}}_2& {\mathrm{m}}_2& {\mathrm{n}}_2\end{array}\right|=0$
Now, 
$\begin{array}{l}\left|\begin{array}{ccc}{\mathrm{x}}_2-{\mathrm{x}}_1& {\mathrm{y}}_2-{\mathrm{y}}_1& {\mathrm{z}}_2-{\mathrm{z}}_1\\ {\mathrm{l}}_1& {\mathrm{m}}_1& {\mathrm{n}}_1\\ {\mathrm{l}}_2& {\mathrm{m}}_2& {\mathrm{n}}_2\end{array}\right|=\left|\begin{array}{ccc}4-1& 1-2& 0-3\\ 2& 3& 4\\ 5& 2& 1\end{array}\right|\\ \Rightarrow \left|\begin{array}{ccc}3& -1& -3\\ 2& 3& 4\\ 5& 2& 1\end{array}\right|\\ =3(3-8)-(-1)(2-20)-3(4-15)\\ =3(-5)-18-3(-11)\\ =-15-18+33=-33+33=0\end{array}$
Therefore, we can say that lines are coplanar. Hence, given lines are not skew.