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Lines \( \frac{x-2}{1}+\frac{y-3}{1}=\frac{z-4}{-K} \) and \( \frac{x-1}{K}=\frac{y-4}{2}=\frac{z-5}{1} \) are co-planar if
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Verified Answer
The correct answer is:
\( \mathrm{K}=0 \)
Given equation,
\[
\begin{array}{l}
\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-K} \rightarrow(1) \\
\frac{x-1}{K}=\frac{y-4}{2}=\frac{z-5}{1} \rightarrow(2)
\end{array}
\]
For co-planarity, we have
\[
\begin{array}{l}
\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{2} & z_{2}-z_{1} \\
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2}
\end{array}\right|=0 \\
\Rightarrow\left|\begin{array}{cccc}
1-2 & 4-3 & 5-4 \\
1 & 1 & -K \\
K & 2 & 1
\end{array}\right|=0 \\
\Rightarrow\left|\begin{array}{cccc}
1 & 1 & 1 & -K \\
K & 2 & 1 & =0
\end{array}\right|=0
\end{array}
\]
\[
\begin{array}{l}
\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-K} \rightarrow(1) \\
\frac{x-1}{K}=\frac{y-4}{2}=\frac{z-5}{1} \rightarrow(2)
\end{array}
\]
For co-planarity, we have
\[
\begin{array}{l}
\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{2} & z_{2}-z_{1} \\
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2}
\end{array}\right|=0 \\
\Rightarrow\left|\begin{array}{cccc}
1-2 & 4-3 & 5-4 \\
1 & 1 & -K \\
K & 2 & 1
\end{array}\right|=0 \\
\Rightarrow\left|\begin{array}{cccc}
1 & 1 & 1 & -K \\
K & 2 & 1 & =0
\end{array}\right|=0
\end{array}
\]
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