If the two 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  each other, then value of k is​

Correct option is B

Given:

​$$\text{Line 1: } \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} \Rightarrow \begin{cases}x = 1 + 2\lambda \\y = -1 + 3\lambda \\z = 1 + 4\lambda\end{cases} \\[6pt]$$

$$\text{Line 2: } \frac{x - 3}{1} = \frac{y - k}{2} = z \Rightarrow \begin{cases}x = 3 + \mu \\y = k + 2\mu \\z = \mu\end{cases} \\[6pt]$$

Solution:

​$$\text{Set equations equal for intersection:} \\1 + 2\lambda = 3 + \mu \tag {1}$$

$$-1 + 3\lambda = k + 2\mu \tag {2}$$

$$1 + 4\lambda = \mu \tag{3}$$

$$\text{From (1) and (3):} \\2\lambda - \mu = 2 \\\mu = 1 + 4\lambda \Rightarrow 2\lambda - (1 + 4\lambda) = 2 \Rightarrow -2\lambda = 3 \Rightarrow \lambda = -\frac{3}{2} \\\mu = 1 + 4(-\frac{3}{2}) = -5 \\[6pt]\text{Substitute into (2):} \\-1 + 3(-\frac{3}{2}) = k + 2(-5) \Rightarrow -\frac{11}{2} = k - 10 \Rightarrow k = \frac{9}{2} \\[6pt]$$

$$\boxed{\text{Final Answer: } k = \frac{9}{2}}$$​​​​​​