Q. 114.8( 4 Votes )

Find the di

Answer :

The equation of the plane passing through (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is given by the following equation:



According to question,


(x1, y1, z1) = (2, 5, –3)


(x2, y2, z2) = (–2, –3, 5)


(x3, y3, z3) = (5, 3, –3)


Putting these values,



(x – 2)(16) + (y – 5)(24) + (z + 3)(32) = 0


2x + 3y + 4z –7 = 0


Distance of 2x + 3y + 4z –7 = 0 from (7, 2, 4)


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


is given by:




p = √29 units


the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, –3), B(–2, –3, 5) and (5, 3, –3) is √29 units.


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