3d coordinates estimation

Question

lets 3 points A,B,C  in the top floor of the room A(1, 1.5,5) and B(2,1,5) and C(2,2,5)  the another point P(x,y,z ) in the room ,the distance between  P-A is 3.301m and P-B is 3.61 m and P-C is 3.62m . Now we have to calculate the position of the point P(x,y,z) ?

Answer 1

Length AP=√((x-1)²+(y-1.5)²+(z-5)²)=3.301m;

length BP=√((x-2)²+(y-1)²+(z-5)²)=3.61m;

length CP=√((x-2)²+(y-2)²+(z-5)²)=3.62m.

BP²-AP²=(x-2+x-1)(x-2-x+1)+(y-1+y-1.5)(y-1-y+1.5)=

-(2x-3)+0.5(2y-2.5)=

-2x+3+y-1.25=-2x+y+1.75.

CP²-BP²=(y-2+y-1)(y-2-y+1)=-(2y-3). -2y+3=CP²-BP², y=½(BP²-CP²+3)=½((-0.01)(7.23)+3)=1.46385.

-2x+y+1.75=(3.61-3.301)(3.61+3.301)=2.135499,

-2x=2.135499-1.46385-1.75=-1.078351, x=0.5391755.

BP²=13.0321=2.13400822+0.2151568225+(z-5)².

z-5=3.268475938, z=8.268475938.

P has the approx coordinates (0.5392,1.4639,8.2685).