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If the straight lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar, then the plane (s) containing these two lines is (are)
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The correct answers are:
$y+z=-1$, $y-z=-1$
Given that lines are coplanar.
$\therefore\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2}
\end{array}\right|=\left|\begin{array}{ccc}
2 & 0 & 0 \\
2 & k & 2 \\
5 & 2 & k
\end{array}\right|=0 \Rightarrow k=\pm 2$
For $k=2$, equation of the plane is given by $\left|\begin{array}{ccc}x-1 & y+1 & z \\ 2 & 2 & 2 \\ 5 & 2 & 2\end{array}\right|=0 \Rightarrow y-z+1=0$
For $k=-2$, equation of the plane is given by $\left|\begin{array}{ccc}x-1 & y+1 & z \\ 2 & -2 & 2 \\ 5 & 2 & -2\end{array}\right|=0 \Rightarrow y+z+1=0$
$\therefore\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2}
\end{array}\right|=\left|\begin{array}{ccc}
2 & 0 & 0 \\
2 & k & 2 \\
5 & 2 & k
\end{array}\right|=0 \Rightarrow k=\pm 2$
For $k=2$, equation of the plane is given by $\left|\begin{array}{ccc}x-1 & y+1 & z \\ 2 & 2 & 2 \\ 5 & 2 & 2\end{array}\right|=0 \Rightarrow y-z+1=0$
For $k=-2$, equation of the plane is given by $\left|\begin{array}{ccc}x-1 & y+1 & z \\ 2 & -2 & 2 \\ 5 & 2 & -2\end{array}\right|=0 \Rightarrow y+z+1=0$
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