If the lines $$\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}$$ and $$\frac{x-3}{1} = \frac{y-k}{2} = \frac{z}{1}$$ intersect, then $$k$$ is equal to

Write each line in parametric form.

For $$\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-1}{4}$$ put the common ratio $$=t$$.
Then $$x=1+2t,\;y=-1+3t,\;z=1+4t$$ $$-(1)$$

For $$\dfrac{x-3}{1}=\dfrac{y-k}{2}=\dfrac{z}{1}$$ put the common ratio $$=s$$.
Then $$x=3+s,\;y=k+2s,\;z=s$$ $$-(2)$$

If the two lines intersect, there must be real numbers $$t$$ and $$s$$ such that the coordinates in $$-(1)$$ and $$-(2)$$ coincide:

$$1+2t = 3+s$$ $$-(3)$$
$$-1+3t = k+2s$$ $$-(4)$$
$$1+4t = s$$ $$-(5)$$

From $$-(5)$$: $$s = 1+4t$$.

Substitute this value of $$s$$ into $$-(3)$$:
$$1+2t = 3 + (1+4t) = 4 + 4t$$
$$1 + 2t - 4t = 4$$
$$-2t = 3 \;\Longrightarrow\; t = -\dfrac{3}{2}$$.

Now find $$s$$ using $$-(5)$$:
$$s = 1 + 4\left(-\dfrac{3}{2}\right) = 1 - 6 = -5$$.

Finally use $$t$$ and $$s$$ in equation $$-(4)$$ to obtain $$k$$:

Left-hand side: $$-1 + 3t = -1 + 3\left(-\dfrac{3}{2}\right) = -1 - \dfrac{9}{2} = -\dfrac{11}{2}$$.

Right-hand side: $$k + 2s = k + 2(-5) = k - 10$$.

Equate the two sides:
$$k - 10 = -\dfrac{11}{2}$$
$$k = -\dfrac{11}{2} + 10 = -\dfrac{11}{2} + \dfrac{20}{2} = \dfrac{9}{2}$$.

Hence $$k = \dfrac{9}{2}$$.

Option C which is: $$\dfrac{9}{2}$$