let x=a+1 a=x-1
y=b+2 b=y-2
z=c+3 c=z-3
because a+b+c=0
so x-1+y-2+z-3=0
therefore x+y+z=6

because 1/a+1 + 1/b+2 + 1/c+3 =0
therefore 1/x + 1/y + 1/z = 0

x+y+z=6
square both sides
(x+y+z)^2=36
expand x^2+y^2+z^2+2xy+2yz+2xz=36
=x^2+y^2+z^2+2(xy+yz+xz)

1/x + 1/y + 1/z = 0
times xyz on both sides
so yz+xy+xz=0
sub. back to x^2+y^2+z^2+2(xy+yz+xz)=36
x^2+y^2+z^2+2(0)=36
therefore x^2+y^2+z^2=36
hencce (a+1)^2+(b+2)^2+(c+3)^2=36 QED.