Step 1: Note the data:
Line 1 passes through $A(1, 2, 3)$ with direction $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
Line 2 passes through $B(4, 1, 0)$ with direction $\vec{b}_2 = 5\hat{i} + 2\hat{j} + \hat{k}$.
Step 2: Parallel or not:
Parallel lines need $\vec{b}_1 = k\vec{b}_2$ for some number $k$.
Here $\frac{2}{5}$, $\frac{3}{2}$ and $\frac{4}{1}$ are all different. So the lines are not parallel.
Step 3: Test for coplanarity:
Two non-parallel lines meet exactly when the shortest distance is zero. This means $(\vec{AB})\cdot(\vec{b}_1 \times \vec{b}_2) = 0$.
\[ \vec{AB} = 3\hat{i} - \hat{j} - 3\hat{k} \]
\[ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{vmatrix} = -5\hat{i} + 18\hat{j} - 11\hat{k} \]
\[ \vec{AB}\cdot(\vec{b}_1 \times \vec{b}_2) = -15 - 18 + 33 = 0 \]
The dot product is zero, so the lines lie in one plane and are intersecting.
Step 4: Find the meeting point:
Take a point $(2\lambda + 1, 3\lambda + 2, 4\lambda + 3)$ on line 1.
For line 2 the ratios give $\frac{x-4}{5} = z$ and $\frac{y-1}{2} = z$. So a point on line 2 has $x = 5z + 4$ and $y = 2z + 1$.
Equate the $z$ values: $z = 4\lambda + 3$. Then $x = 20\lambda + 19$ and $y = 8\lambda + 7$.
Match with line 1: $2\lambda + 1 = 20\lambda + 19$ gives $\lambda = -1$.
Check the $y$ value: $3(-1) + 2 = -1$ and $8(-1) + 7 = -1$. They match.
The point is $(-1, -1, -1)$.
Final Answer:
The lines are not parallel, and they meet at the point below. \[ \boxed{(-1, -1, -1)} \]