Find an equation for the set of all points that are equidistant from the planes 3x − 4y + 12z = 6 and 4x + 3z = 7.

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Solution

$Let (x, y) be a point which is equidistant from the given planes. Then, \frac{|3 x - 4 y + 12 z - 6|}{\sqrt{9 + 16 + 144}} = \frac{|4 x + 3 z - 7|}{\sqrt{16 + 9}} \Rightarrow \frac{3 x - 4 y + 12 z - 6}{13} = \pm \frac{4 x + 3 z - 7}{5} \Rightarrow 15 x - 20 y + 60 z - 30 = 52 x + 39 z - 91; 15 x - 20 y + 60 z - 30 = - 52 x - 39 z + 91 \Rightarrow 37 x + 20 y - 21 z - 61 = 0; 67 x - 20 y + 99 z = 121$

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