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Question
Show that the lines \[\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1} \text{ and }\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}\] are coplanar. Also, find the equation of the plane containing them.
Solution
\[\text{ We know that the lines } \]
\[\frac{x - x_1}{l_1} = \frac{y - y_1}{m_1} = \frac{z - z_1}{n_1} \text{ and } \frac{x - x_2}{l_2} = \frac{y - y_2}{m_2} = \frac{z - z_2}{n_2} \text{ are coplanar if } \]
\[\begin{vmatrix}x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{vmatrix} = 0\]
\[\text{ and the equation of the plane containing these lines is } \]
\[\begin{vmatrix}x - x_1 & y - y_1 & z - z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{vmatrix} = 0\]
\[\text{ Here } ,\]
\[ x_1 = - 1; y_1 = 3; z_1 = - 2; x_2 = 0; y_2 = 7; z_2 = - 7; l_1 = - 3; m_1 = 2; n_1 = 1; l_2 = 1; m_2 = - 3; n_2 = 2\]
\[\text{ Now } ,\]
\[\begin{vmatrix}0 + 1 & 7 - 3 & - 7 + 2 \\ - 3 & 2 & 1 \\ 1 & - 3 & 2\end{vmatrix}\]
\[ = \begin{vmatrix}1 & 4 & - 5 \\ - 3 & 2 & 1 \\ 1 & - 3 & 2\end{vmatrix}\]
\[ = 1 \left( 7 \right) - 4 \left( - 7 \right) - 5 \left( 7 \right)\]
\[ = 7 + 28 - 35\]
\[ = 0\]
\[\text{ So, the given lines are coplanar } .\]
\[\text{ The equation of the plane containing the given lines is } \]
\[\begin{vmatrix}x + 1 & y - 3 & z + 2 \\ - 3 & 2 & 1 \\ 1 & - 3 & 2\end{vmatrix} = 0\]
\[ \Rightarrow \left( x + 1 \right) \left( 7 \right) - \left( y - 3 \right) \left( - 7 \right) + \left( z + 2 \right) \left( 7 \right) = 0\]
\[ \Rightarrow 7x + 7y + 7z = 0\]
\[ \Rightarrow x + y + z = 0\]
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