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Question
Find an equation for the set of all points that are equidistant from the planes 3x − 4y + 12z = 6 and 4x + 3z = 7.
Solution
\[ \text{ Let } \left( x, y \right) \text{ be a point which is equidistant from the given planes. Then } ,\]
\[\frac{\left| 3x - 4y + 12z - 6 \right|}{\sqrt{9 + 16 + 144}} = \frac{\left| 4x + 3z - 7 \right|}{\sqrt{16 + 9}}\]
\[ \Rightarrow \frac{3x - 4y + 12z - 6}{13} = \pm \frac{4x + 3z - 7}{5}\]
\[ \Rightarrow 15x - 20y + 60z - 30 = 52x + 39z - 91; 15x - 20y + 60z - 30 = - 52x - 39z + 91\]
\[ \Rightarrow 37x + 20y - 21z - 61 = 0; 67x - 20y + 99z = 121\]
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