Question

Find the equation of the plane passing through the line of intersection of the planes 2x − 7y + 4z − 3 = 0, 3x − 5y + 4z + 11 = 0 and the point (−2, 1, 3).

Answer 1

\[\text{ The equation of the plane passing through the line of intersection of the given planes is } \]

\[2x - 7y + 4z - 3 + \lambda \left( 3x - 5y + 4z + 11 \right) = 0 . . . \left( 1 \right)\]

\[\text{ This passes through (-2, 1, 3). So } ,\]

\[ - 4 - 7 + 12 - 3 + \lambda \left( - 6 - 5 + 12 + 11 \right) = 0\]

\[ \Rightarrow - 2 + 12\lambda = 0\]

\[ \Rightarrow \lambda = \frac{1}{6}\]

\[\text{ Substituting this in (1), we get } \]

\[2x - 7y + 4z - 3 + \frac{1}{6} \left( 3x - 5y + 4z + 11 \right) = 0\]

\[ \Rightarrow 12x - 42y + 24z - 18 + 3x - 5y + 4z + 11 = 0\]

\[ \Rightarrow 15x - 47y + 28z = 7\]