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Question
Find the equation of the plane passing through the points (1, −1, 2) and (2, −2, 2) and which is perpendicular to the plane 6x − 2y + 2z = 9.
Solution
\[ \text{ The equation of any plane passing through (1, -1, 2) is } \]
\[a \left( x - 1 \right) + b \left( y + 1 \right) + c \left( z - 2 \right) = 0 . . . \left( 1 \right)\]
\[ \text{ It is given that (1) is passing through (2, -2, 2). So, }\]
\[a \left( 2 - 1 \right) + b \left( - 2 + 1 \right) + c \left( 2 - 2 \right) = 0 \]
\[ \Rightarrow a - b + 0c = 0 . . . \left( 2 \right)\]
\[ \text{ It is given that (1) is perpendicular to the plane 6x - 2y + 2z = 9 . So,} \]
\[6a - 2b + 2c = 0\]
\[ \Rightarrow 3a - b + c = 0 . . . \left( 3 \right)\]
\[\text{ Solving (1), (2) and (3), we get } \]
\[\begin{vmatrix}x - 1 & y + 1 & z - 2 \\ 1 & - 1 & 0 \\ 3 & - 1 & 1\end{vmatrix} = 0\]
\[ \Rightarrow - 1 \left( x - 1 \right) - 1 \left( y + 1 \right) + 2 \left( z - 2 \right) = 0\]
\[ \Rightarrow - x + 1 - y - 1 + 2z - 4 = 0\]
\[ \Rightarrow x + y - 2z + 4 = 0\]
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