JEE MAIN - Mathematics (2013 (Offline) - No. 21)
If 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, then $$k$$ can have :
any value
exactly one value
exactly two values
exactly three values
Explanation
Given lines will be coplanar
If $$\,\,\,\,\left| {\matrix{ { - 1} & 1 & 1 \cr 1 & 1 & { - k} \cr k & 2 & 1 \cr } } \right| = 0$$
$$ \Rightarrow - 1\left( {1 + 2k} \right) - \left( {1 + {k^2}} \right) + 1\left( {2 - k} \right) = 0$$
$$ \Rightarrow k = 0, - 3$$
If $$\,\,\,\,\left| {\matrix{ { - 1} & 1 & 1 \cr 1 & 1 & { - k} \cr k & 2 & 1 \cr } } \right| = 0$$
$$ \Rightarrow - 1\left( {1 + 2k} \right) - \left( {1 + {k^2}} \right) + 1\left( {2 - k} \right) = 0$$
$$ \Rightarrow k = 0, - 3$$
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