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Question

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


A
k=1 or k=1
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B
k=0 or k=3
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C
k=3 or k=3
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D
k=0 or k=1
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Solution

The correct option is A $$k=0$$ or $$k=-3$$
Using fact, two line $$\displaystyle \frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$$ and  $$\displaystyle \frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$$ are coplanar if
$$\displaystyle \begin{vmatrix}x_{2}-x_{1} &y_{2}-y_{1}  &z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1}\\ a_{2} &b_{2}  &c_{2} \end{vmatrix}=0$$
$$\Rightarrow \displaystyle \begin{vmatrix}-1 &1  &1 \\  1& 1 &-k \\  k& 2 &1 \end{vmatrix}=0 \Rightarrow k^{2}+3k=0$$
$$\Rightarrow k=0$$ or $$k=-3$$

Mathematics

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