The line and x−21=y−31=z−4−k and x−1k=y−42=z−51 are coplanar if

The lines x−21=y−31=z−4−k -------(i)and x−1k=y−42=z−51-------(ii)are coplanar if x2−x1y2−y1z2−z1l1m1n1l2m1n2=0or 1−1−111−kk21=0or k2+3k=0⇒k=0 or −3

The line and x−21=y−31=z−4−k and x−1k=y−42=z−51 are coplanar if

The lines x−21=y−31=z−4−k -------(i)and x−1k=y−42=z−51-------(ii)are coplanar if x2−x1y2−y1z2−z1l1m1n1l2m1n2=0or 1−1−111−kk21=0or k2+3k=0⇒k=0 or −3

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The line and $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar if

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k=1 or -1

k=0 or -3

k=3 or -3

k=0 or -1

answer is B.

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Detailed Solution

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ -------(i)

and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ -------(ii)

are coplanar if $|\begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{1} & n_{2} \end{array}| = 0$

$\begin{array}{l} or |\begin{array}{ccc} 1 & - 1 & - 1 \\ 1 & 1 & - k \\ k & 2 & 1 \end{array}| = 0 \\ or k^{2} + 3 k = 0 \\ \Rightarrow k = 0 or - 3 \end{array}$