Question
The lines $$\frac{{x - 2}}{1} = \frac{{y - 3}}{1} = \frac{{z - 4}}{{ - k}}$$ and $$\frac{{x - 1}}{k} = \frac{{y - 4}}{1} = \frac{{z - 5}}{1}$$ are coplanar if :
A.
$$k=3$$ or $$-2$$
B.
$$k=0$$ or $$-1$$
C.
$$k=1$$ or $$-1$$
D.
$$k=0$$ or $$-3$$
Answer :
$$k=0$$ or $$-3$$
Solution :
\[\begin{array}{l}
\left| \begin{array}{l}
{x_2} - {x_1}\,\,\,\,\,{y_2} - {y_1}\,\,\,\,\,{z_2} - {z_1}\\
\,\,\,\,\,\,\,\,\,{l_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{m_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{n_1}\\
\,\,\,\,\,\,\,\,{l_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{m_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{n_2}
\end{array} \right| = 0\\
\therefore \left| \begin{array}{l}
1\,\,\,\,\, - 1\,\,\,\,\, - 1\\
1\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\, - k\\
k\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,1
\end{array} \right| = 0\,\,\,\,\, \Rightarrow \left| \begin{array}{l}
\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\, - 1\\
\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,1 + k\,\,\,\,\, - k\\
k + 2\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1
\end{array} \right| = 0\\
{k^2} + 3k = 0\\
\Rightarrow k\left( {k + 3} \right) = 0\\
\Rightarrow k = 0\,\,{\rm{ or }} - 3
\end{array}\]