Step 1: Direction vectors:
Let $\vec{b}_1 = \overrightarrow{AB} = 4\hat{i} + 6\hat{j} + 2\hat{k}$ and $\vec{b}_2 = \overrightarrow{CD} = -7\hat{i} - 5\hat{j}$.
These are not scalar multiples of each other, so the lines are not parallel.
Step 2: Use the coplanarity test:
Two non-parallel lines meet exactly when the vector between a point on each is perpendicular to $\vec{b}_1 \times \vec{b}_2$.
\[ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 6 & 2 \\ -7 & -5 & 0 \end{vmatrix} = 10\hat{i} - 14\hat{j} + 22\hat{k} \]
Step 3: Find the dot product:
$\overrightarrow{AC} = (3 - 0,\ 9 + 1,\ 4 + 1) = (3, 10, 5)$.
\[ \overrightarrow{AC}\cdot(\vec{b}_1\times\vec{b}_2) = 3(10) + 10(-14) + 5(22) = 30 - 140 + 110 = 0 \]
The value is zero, so the lines are coplanar. As they are not parallel, they intersect.
Step 4: Locate the common point:
Every point of CD has $z = 4$.
Line AB in symmetric form is $\dfrac{x}{2} = \dfrac{y+1}{3} = \dfrac{z+1}{1}$.
Putting $z = 4$ gives $\dfrac{z+1}{1} = 5$, so $x = 10$ and $y + 1 = 15$, that is $y = 14$.
The point $P(10, 14, 4)$ lies on AB.
Step 5: Confirm P lies on CD:
Line CD is $\dfrac{x - 3}{-7} = \dfrac{y - 9}{-5},\ z = 4$.
At $P$: $\dfrac{10 - 3}{-7} = -1$ and $\dfrac{14 - 9}{-5} = -1$. The two ratios match.
So $P$ is the point of intersection.
Step 6: Line OP:
The line joins $O(0,0,0)$ and $P(10,14,4)$, so its direction ratios are $10 : 14 : 4 = 5 : 7 : 2$.
\[ \vec{r} = \lambda\,(5\hat{i} + 7\hat{j} + 2\hat{k}) \]
Final Answer:
The lines meet at $(10, 14, 4)$ and the line through the origin is $\dfrac{x}{5} = \dfrac{y}{7} = \dfrac{z}{2}$. \[ \boxed{\vec{r} = \lambda\,(5\hat{i} + 7\hat{j} + 2\hat{k})} \]