Question

Find the equation of the plane that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8.

Answer 1

\[ \text{ The equation of any plane passing through (1, -1, 2) is } \]

\[a \left( x - 1 \right) + b \left( y + 1 \right) + c \left( z - 2 \right) = 0 . . . \left( 1 \right)\]

\[ \text{ It is given that (1) is perpendicular to the plane 2x + 3y - 2z = 5 . So, } \]

\[2a + 3b - 2c = 0 . . . \left( 2 \right)\]

\[\text{ It is given that (1) is perpendicular to the plane x + 2y - 3z = 8 . So } ,\]

\[a + 2b - 3c = 0 . . . \left( 3 \right)\]

\[\text{ Solving (1), (2) and (3), we get } \]

\[\begin{vmatrix}x - 1 & y + 1 & z - 2 \\ 2 & 3 & - 2 \\ 1 & 2 & - 3\end{vmatrix} = 0\]

\[ \Rightarrow - 5 \left( x - 1 \right) + 4 \left( y + 1 \right) + 1 \left( z - 2 \right) = 0\]

\[ \Rightarrow 5x - 4y - z = 7\]