Given that x1,y1,z1 are in AP and x,2y,z are in GP.
As given, y2=x1+y1.......(i)
Also, 2y2=xz.......(ii)
Also given that xy+yz+zx=23xyz
⇒x1+y1+z1=23.....(iii)
From (i) and (iii) we get y3=23
y=2.....(iv)
Now from (ii) xz=4.......(v)
Now using (ii),(iv)and(v)
⇒x+z=42
Hence 3(x+y+z)2=3(2+42)2
=150
Therefore, this is the required answer.