Shortest distance between the two given lines is given as:
$$
d=\frac{\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|}{\sqrt{\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}}
$$
$$
\Rightarrow d=\frac{\left|\begin{array}{ccc}
4-1 & 1-2 & 0-3 \\
2 & 3 & 4 \\
5 & 2 & 1
\end{array}\right|}{\sqrt{(3 \cdot 1-2 \cdot 4)^2+(4 \cdot 5-1 \cdot 2)^2+(2 \cdot 2-5 \cdot 3)^2}}
$$
$\Rightarrow d=\frac{\left|\begin{array}{ccc}3 & -1 & -3 \\ 2 & 3 & 4 \\ 5 & 2 & 1\end{array}\right|}{\sqrt{25+324+121}}$


$$
\Rightarrow \mathrm{d}=\frac{3(3-8)+1(2-20)-3(4-15)}{\sqrt{470}}
$$
$$
\Rightarrow \mathrm{d}=\frac{-15-18+33}{\sqrt{470}}=\frac{0}{\sqrt{470}}=0
$$
Since, the shortest distance between the two given lines is zero. Therefore, the given two lines are intersecting.
For finding their point of intersection for first line.
$$
\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda
$$
$$
\Rightarrow \mathrm{x}=2 \lambda+1, \mathrm{y}=3 \lambda+2
$$
and $\mathrm{z}=4 \lambda+3$
Since, the lines are intersecting. So, let us put these values in the equation of another line.
Thus, $\frac{2 \lambda+1-4}{5}=\frac{3 \lambda+2-1}{2}=\frac{4 \lambda+3}{1}$

$$
\begin{aligned}
&\Rightarrow \quad \frac{2 \lambda-3}{5}=\frac{3 \lambda+1}{2}=\frac{4 \lambda+3}{1} \Rightarrow \frac{2 \lambda-3}{5}=\frac{4 \lambda+3}{1} \\
&\Rightarrow \quad 2 \lambda-3=20 \lambda+15 \Rightarrow 18 \lambda=-18 \Rightarrow \lambda=-1
\end{aligned}
$$
So, the required point of intersection is
$$
\begin{aligned}
&x=2(-1)+1=-1 \\
&y=3(-1)+2=-1 \\
&z=4(-1)+3=-1
\end{aligned}
$$
Thus, the given lines intersect at $(-1,-1,-1)$.