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

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(D) Let the given lines be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu$.Any point on the firs

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$\frac{9}{2}$

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(D) Let the given lines be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu$.Any point on the first line is $(2\lambda+1, 3\lambda-1, 4\lambda+1)$ and any point on the second line is $(\mu+3, 2\mu+k, \mu)$.Since the lines intersect,there must exist some $\lambda$ and $\mu$ such that the coordinates are equal:$2\lambda+1 = \mu+3 \implies 2\lambda - \mu = 2$  .... $(1)$$3\lambda-1 = 2\mu+k \implies 3\lambda - 2\mu = k+1$  .... $(2)$$4\lambda+1 = \mu \implies 4\lambda - \mu = -1$  .... $(3)$Subtracting $(1)$ from $(3)$,we get $(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2$,which gives $2\lambda = -3$,so $\lambda = \frac{-3}{2}$.Substituting $\lambda = \frac{-3}{2}$ into $(3)$,we get $\mu = 4(\frac{-3}{2}) + 1 = -6 + 1 = -5$.Now,substitute $\lambda = \frac{-3}{2}$ and $\mu = -5$ into $(2)$:$3(\frac{-3}{2}) - 2(-5) = k+1$$\frac{-9}{2} + 10 = k+1$$\frac{-9+20}{2} = k+1$$\frac{11}{2} = k+1$$k = \frac{11}{2} - 1 = \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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