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Question
Find the equations of the planes parallel to the plane x + 2y − 2z + 8 = 0 that are at a distance of 2 units from the point (2, 1, 1).
Solution
\[ \text{ The equation of the plane parallel to the given plane is } \]
\[x + 2y - 2z + k = 0 . . . \left( 1 \right)\]
\[ \text{ It is given that plane (1) is at a distance of 2 units from (2, 1, 1) } .\]
\[ \Rightarrow \frac{\left| 2 + 2 - 2 + k \right|}{\sqrt{1^2 + 2^2 + \left( - 2 \right)^2}} = 2\]
\[ \Rightarrow \frac{\left| 2 + k \right|}{3} = 2\]
\[ \Rightarrow \left| 2 + k \right| = 6\]
\[ \Rightarrow 2 + k = 6; 2 + k = - 6\]
\[ \Rightarrow k = 4; k = - 8\]
\[ \text { Substituting these two values one by one in (1), we get } \]
\[x + 2y - 2z + 4 = 0 \text{ and }\]
\[x + 2y - 2z - 8 = 0, \text{ which are the equations of the required planes } .\]
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