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

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(C) Let the two lines be represented by parameters $\lambda$ and $\mu$ respectively: $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda \implies x=2\l

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$\frac{9}{2}$

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(C) Let the two lines be represented by parameters $\lambda$ and $\mu$ respectively: $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda \implies x=2\lambda+1, y=3\lambda-1, z=4\lambda+1$ $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu \implies x=\mu+3, y=2\mu+k, z=\mu$ If the lines intersect,there must exist a point common to both,so: $2\lambda+1 = \mu+3 \implies 2\lambda - \mu = 2 \quad (i)$ $3\lambda-1 = 2\mu+k \implies 3\lambda - 2\mu = k+1 \quad (ii)$ $4\lambda+1 = \mu \implies 4\lambda - \mu = -1 \quad (iii)$ Subtracting $(i)$ from $(iii)$: $(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2 \implies 2\lambda = -3 \implies \lambda = -\frac{3}{2}$ Substituting $\lambda = -\frac{3}{2}$ into $(iii)$: $4(-\frac{3}{2}) + 1 = \mu \implies -6 + 1 = \mu \implies \mu = -5$ Substituting $\lambda = -\frac{3}{2}$ 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 $k$ is equal to:

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