. If the Iines x-12=y+13=z-14 and x-31=y-k2=z1intersect, then the value of k is

Then, given lines will intersect, if3-1k+10-1234121=0⇒2(3-8)-(k+1)(2-4)-(4-3)=0⇒k=92

. If the Iines x-12=y+13=z-14 and x-31=y-k2=z1intersect, then the value of k is

Then, given lines will intersect, if3-1k+10-1234121=0⇒2(3-8)-(k+1)(2-4)-(4-3)=0⇒k=92

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. If the Iines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} and \frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$
intersect, then the value of k is

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$\frac{3}{2}$

$\frac{9}{2}$

$- \frac{2}{9}$

$- \frac{3}{2}$

answer is B.

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Then, given lines will intersect, if

$|\begin{matrix} 3 - 1 & k + 1 & 0 - 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{matrix}| = 0$

$\Rightarrow 2 (3 - 8) - (k + 1) (2 - 4) - (4 - 3) = 0 \Rightarrow k = \frac{9}{2}$

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. If the Iines x-12=y+13=z-14 and x-31=y-k2=z1intersect, then the value of k is

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. If the Iines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} and \frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$
intersect, then the value of k is