JEE MAIN - Mathematics (2003 - No. 29)
The lines $${{x - 2} \over 1} = {{y - 3} \over 1} = {{z - 4} \over { - k}}$$ and $${{x - 1} \over k} = {{y - 4} \over 2} = {{z - 5} \over 1}$$ are coplanar if :
$$k=3$$ or $$-2$$
$$k=0$$ or $$-1$$
$$k=1$$ or $$-1$$
$$k=0$$ or $$-3$$
Explanation
Coplanar if
$$\left| {\matrix{ {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr {{l_1}} & {{m_1}} & {{n_1}} \cr {{l_2}} & {{m_2}} & {{n_2}} \cr } } \right| = 0$$
$$\therefore$$ $$\left| {\matrix{ 1 & { - 1} & { - 1} \cr 1 & 1 & { - k} \cr k & 2 & 1 \cr } } \right| = 0$$
$$ \Rightarrow \left| {\matrix{ 0 & 0 & { - 1} \cr 2 & {1 + k} & { - k} \cr {k + 2} & 1 & 1 \cr } } \right| = 0$$
$${k^2} + 3k = 0 \Rightarrow k\left( {k + 3} \right) = 0$$
or $$k = 0$$ or $$-3$$
$$\left| {\matrix{ {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr {{l_1}} & {{m_1}} & {{n_1}} \cr {{l_2}} & {{m_2}} & {{n_2}} \cr } } \right| = 0$$
$$\therefore$$ $$\left| {\matrix{ 1 & { - 1} & { - 1} \cr 1 & 1 & { - k} \cr k & 2 & 1 \cr } } \right| = 0$$
$$ \Rightarrow \left| {\matrix{ 0 & 0 & { - 1} \cr 2 & {1 + k} & { - k} \cr {k + 2} & 1 & 1 \cr } } \right| = 0$$
$${k^2} + 3k = 0 \Rightarrow k\left( {k + 3} \right) = 0$$
or $$k = 0$$ or $$-3$$
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