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