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

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(A) The given lines are $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = \lambda$ and $L_2: \frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1} = \mu$. Any point

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$\frac{9}{2}$

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(A) The given lines are $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = \lambda$ and $L_2: \frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1} = \mu$. Any point on $L_1$ is $P(2\lambda+1, 3\lambda-1, 4\lambda+1)$ and any point on $L_2$ is $Q(\mu+3, 2\mu+k, \mu)$. For the lines to intersect,there must exist $\lambda$ and $\mu$ such that $P=Q$. Equating coordinates: $2\lambda+1 = \mu+3 \implies 2\lambda - \mu = 2$ $(i)$ $3\lambda-1 = 2\mu+k \implies 3\lambda - 2\mu = k+1$ (ii) $4\lambda+1 = \mu \implies 4\lambda - \mu = -1$ (iii) Subtracting $(i)$ from (iii): $(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2 \implies 2\lambda = -3 \implies \lambda = -\frac{3}{2}$. Substitute $\lambda = -\frac{3}{2}$ into $(i)$: $2(-\frac{3}{2}) - \mu = 2 \implies -3 - \mu = 2 \implies \mu = -5$. Now substitute $\lambda = -\frac{3}{2}$ and $\mu = -5$ into (ii): $3(-\frac{3}{2}) - 2(-5) = k+1 \implies -\frac{9}{2} + 10 = k+1 \implies k = 9 - \frac{9}{2} = \frac{9}{2}$.

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 $k=$

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