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

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(B) Any point on the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ is given by $(2\lambda+1, 3\lambda-1, 4\lambda+1)$.Any point on the line

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$\frac{9}{2}$

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(B) Any point on the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ is given by $(2\lambda+1, 3\lambda-1, 4\lambda+1)$.Any point on the line $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu$ is given by $(\mu+3, 2\mu+k, \mu)$.For the lines to intersect,there must exist values of $\lambda$ and $\mu$ such that:$2\lambda+1 = \mu+3$  $(i)$$3\lambda-1 = 2\mu+k$  $(ii)$$4\lambda+1 = \mu$  $(iii)$Substituting $\mu = 4\lambda+1$ from $(iii)$ into $(i)$:$2\lambda+1 = (4\lambda+1)+3$$2\lambda+1 = 4\lambda+4$$-2\lambda = 3 \Rightarrow \lambda = -\frac{3}{2}$.Now,find $\mu$ using $(iii)$:$\mu = 4(-\frac{3}{2})+1 = -6+1 = -5$.Substitute $\lambda = -\frac{3}{2}$ and $\mu = -5$ into $(ii)$:$3(-\frac{3}{2})-1 = 2(-5)+k$$-\frac{9}{2}-1 = -10+k$$-\frac{11}{2} = -10+k$$k = 10 - \frac{11}{2} = \frac{20-11}{2} = \frac{9}{2}$.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

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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