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

Let the set of points be denoted by (x1,y1,z1)


Distance of (x1,y1,z1) from 3x – 4y + 12z = 6


We know, the distance of point (x1,y1,z1) from the plane


is given by:





Similarly,


Distance of (x1,y1,z1) from 4x + 3z = 7:




According to the question,




or


37x1 + 20y1 – 21z1 –61 = 0 or 67x1 –20 y1 + 99z1 –121 = 0


Equations of set of points equidistant from planes 3x – 4y + 12z = 6 and 4x + 3z = 7 is 37x1 + 20y1 – 21z1 –61 = 0 or 67x1 –20 y1 + 99z1 – 121 = 0


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