If the lines frac{x-1}{2}=frac{y+1}{3}=frac{z | vedclass

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(C) Let the first line be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = \lambda$. Then any point on this line is $(2\lambda+1, 3\lambda-1, 4\lambda+1)$

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$\frac{9}{2}$

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(C) Let the first line be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = \lambda$. Then any point on this line is $(2\lambda+1, 3\lambda-1, 4\lambda+1)$.Let the second line be $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1} = \mu$. Then any point on this line is $(\mu+3, 2\mu+k, \mu)$.If the lines intersect,there exist $\lambda$ and $\mu$ such that:$2\lambda+1 = \mu+3 \implies 2\lambda - \mu = 2$ (Equation $1$)$4\lambda+1 = \mu \implies 4\lambda - \mu = -1$ (Equation $2$)Subtracting Equation $1$ from Equation $2$: $(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2 \implies 2\lambda = -3 \implies \lambda = -\frac{3}{2}$.Substituting $\lambda = -\frac{3}{2}$ into Equation $1$: $2(-\frac{3}{2}) - \mu = 2 \implies -3 - \mu = 2 \implies \mu = -5$.Now,equate the $y$-coordinates: $3\lambda - 1 = 2\mu + k$.Substitute $\lambda = -\frac{3}{2}$ and $\mu = -5$: $3(-\frac{3}{2}) - 1 = 2(-5) + k$.$-\frac{9}{2} - 1 = -10 + k \implies -\frac{11}{2} = -10 + k$.$k = 10 - \frac{11}{2} = \frac{20-11}{2} = \frac{9}{2}$.

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 the value of $k$ is

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