Let 0<z<y<x be three real numbers such that 1 x, 1 y, 1 z are in an arithmetic progression and x,√2y,z are in a geometric progression. If xy+yz+zx= 3…

Given that $\frac{1}{x},\frac{1}{y},\frac{1}{z}$x1​,y1​,z1​ are in AP and $x,\sqrt{2}y,z$x,2​y,z are in GP.

As given, $\frac{2}{y}=\frac{1}{x}+\frac{1}{y}.......(i)$y2​=x1​+y1​.......(i)

Also, $2{y}^{2}=xz.......(\mathrm{ii})$2y2=xz.......(ii)

Also given that $xy+yz+zx=\frac{3}{\sqrt{2}}xyz$xy+yz+zx=2​3​xyz

$$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{\sqrt{2}}.....(\mathrm{iii})$$⇒x1​+y1​+z1​=2​3​.....(iii)

From $(i)$(i) and $(iii)$(iii) we get $\frac{3}{y}=\frac{3}{\sqrt{2}}$y3​=2​3​

$$y=\sqrt{2}.....(\mathrm{iv})$$y=2​.....(iv)

Now from $(ii)$(ii) $xz=4.......(v)$xz=4.......(v)

Now using $(ii),(iv)\text{and}(v)$(ii),(iv)and(v)

$$\Rightarrow x+z=4\sqrt{2}$$⇒x+z=42​

Hence $3(x+y+z{)}^{2}=3(\sqrt{2}+4\sqrt{2}{)}^{2}$3(x+y+z)2=3(2​+42​)2

$$=150$$=150

Therefore, this is the required answer.