Two lines $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-1}{4}$ and $\dfrac{x-3}{1}=\dfrac{y-k}{2}=\dfrac{z}{1}$ intersect at a point. Then the value of ' $k$ ' is

Let the first line be $L_1: \dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{4} = \lambda$.

Any point on $L_1$ is $P(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$.

Let the second line be $L_2: \dfrac{x-3}{1} = \dfrac{y-k}{2} = \dfrac{z}{1} = \mu$.

Any point on $L_2$ is $Q(\mu + 3, 2\mu + k, \mu)$.

Since the lines intersect, there exist $\lambda$ and $\mu$ such that $P = Q$.

Equating the coordinates:

1) $2\lambda + 1 = \mu + 3 \Rightarrow 2\lambda - \mu = 2$

2) $3\lambda - 1 = 2\mu + k \Rightarrow 3\lambda - 2\mu = k + 1$

3) $4\lambda + 1 = \mu \Rightarrow 4\lambda - \mu = -1$

Subtracting equation (1) from equation (3):

$(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2$

$2\lambda = -3 \Rightarrow \lambda = -\dfrac{3}{2}$

Substituting $\lambda = -\dfrac{3}{2}$ into equation (1):

$2(-\dfrac{3}{2}) - \mu = 2 \Rightarrow -3 - \mu = 2 \Rightarrow \mu = -5$

Substituting $\lambda = -\dfrac{3}{2}$ and $\mu = -5$ into equation (2):

$3(-\dfrac{3}{2}) - 2(-5) = k + 1$

$-\dfrac{9}{2} + 10 = k + 1$

$\dfrac{11}{2} = k + 1$

$k = \dfrac{11}{2} - 1 = \dfrac{9}{2}$

Answer: $\dfrac{9}{2}$