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Question
Obtain the equation of the plane passing through the point (1, −3, −2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.
Solution
\[\text{ The equation of any plane passing through (1, -3, -2) is } \]
\[a \left( x - 1 \right) + b \left( y + 3 \right) + c \left( z + 2 \right) = 0 . . . \left( 1 \right)\]
\[\text{ It is given that (1) is perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8 . Then } ,\]
\[a + 2b + 2c = 0 . . . \left( 2 \right)\]
\[3a + 3b + 2c = 0 . . . \left( 3 \right)\]
\[\text{ Solving (1), (2) and (3), we get } \]
\[\begin{vmatrix}x - 1 & y + 3 & z + 2 \\ 1 & 2 & 2 \\ 3 & 3 & 2\end{vmatrix} = 0\]
\[ \Rightarrow - 2 \left( x - 1 \right) + 4 \left( y + 3 \right) - 3 \left( z + 2 \right) = 0\]
\[ \Rightarrow - 2x + 2 + 4y + 12 - 3z - 6 = 0\]
\[ \Rightarrow 2x - 4y + 3z - 8 = 0\]
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