Melody

 |z - 3| = |z + i| = |z - 1 - 2i|.

let z= a+bi

 \(|a+bi - 3| = |a+bi + i| = |a+bi - 1 - 2i| \\  |(a-3)+bi| = |a+(b+1)i| = | (a -1)+(b - 2)i |\\ (a-3)^2+b^2 = a^2+(b+1)^2 = (a-1)^2+(b-2)^2\\ a^2-6a+9+b^2=a^2+b^2+2b+1=a^2-2a+1+b^2-4b+4\\ -6a+9=2b+1=-2a+1-4b+4\\ -6a+8=2b=-2a-4b+4\\ -3a+4=b=-a-2b+2\\~\\ b=-3a+4\\so\\ -3a+4=-a-2(-3a+4)+2\\ -3a+4=-a+6a-8+2\\ -3a+4=5a-6\\ -8a=-10\\ a=1.25\\ b=-3a+4=-3.75+4=0.25\\ z=\frac{5+i}{4} \)

Latex:

|a+bi - 3| = |a+bi + i| = |a+bi - 1 - 2i| \\
 |(a-3)+bi| = |a+(b+1)i| = | (a -1)+(b - 2)i |\\

(a-3)^2+b^2 = a^2+(b+1)^2 = (a-1)^2+(b-2)^2\\
a^2-6a+9+b^2=a^2+b^2+2b+1=a^2-2a+1+b^2-4b+4\\
-6a+9=2b+1=-2a+1-4b+4\\
-6a+8=2b=-2a-4b+4\\
-3a+4=b=-a-2b+2\\~\\
b=-3a+4\\so\\
-3a+4=-a-2(-3a+4)+2\\
-3a+4=-a+6a-8+2\\
-3a+4=5a-6\\
-8a=-10\\
a=1.25\\
b=-3a+4=-3.75+4=0.25\\
z=\frac{5+i}{4}

I answered but i have decided to hide my answer as this is obviously a school competition homework question.

Thanks Ginger, 

... I think not, and hope not, too :)

Here is the relevent part of the final pic.

You can finish it yourseldf now

Hint: use cosine rule.

It will be of the form 

y=a(x+2)(x+3)

plug in (5,10) to find a

You forgot the sqrt on the bottom EP, just a trivial mistake that SpaceX should have picke  up for himeself.

\(y=\frac{k}{\sqrt x}\\ (2,4)\\ 4=\frac{k}{\sqrt 2}\\ k=4\sqrt 2\\ y=\frac{4\sqrt 2}{\sqrt x}\\ when\;y=1\\ 1=\frac{4\sqrt 2}{\sqrt x}\\ 1=\frac{32}{x}\\ x=32\)