1. Write both lines in parametric form

For the first line, $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=t$ So, $x=1+2t,\quad y=-1+3t,\quad z=1+4t$

For the second line, $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=s$ So, $x=3+s,\quad y=k+2s,\quad z=s$

2. Use the condition that the lines intersect

At the point of intersection, coordinates must be equal for some $t$ and $s$: $1+2t=3+s \quad ...(1)$ $-1+3t=k+2s \quad ...(2)$ $1+4t=s \quad ...(3)$

3. Solve for $t$ and $s$ using (1) and (3)

From (3), $s=1+4t$ Substitute into (1): $1+2t=3+(1+4t)$ $1+2t=4+4t$ $-3=2t$ $t=-\frac{3}{2}$

Now from (3), $s=1+4\left(-\frac{3}{2}\right)=1-6=-5$

4. Find $k$ using equation (2)

Substitute $t=-\frac{3}{2}$ and $s=-5$ into (2): $-1+3\left(-\frac{3}{2}\right)=k+2(-5)$ $-1-\frac{9}{2}=k-10$ $-\frac{11}{2}=k-10$ $k=10-\frac{11}{2}=\frac{20-11}{2}=\frac{9}{2}$

5. Match with the options

$k=\frac{9}{2}$ So the correct option is C.

6. Compare with stored correct answer

Stored correct answer: C

Our derived answer is also C.