11. Let $x+y=a, y+z=b, z+x=c$. Then
$$
a+b>c, a+c>b, b+c>a \text {. }
$$

Using $a, b, c$ as the sides of $\triangle ABC$.
Let $p=\frac{1}{2}(a+b+c)$, combining with Heron's formula we get
$S=\sqrt{p(p-a)(p-b)(p-c)}$
$=\sqrt{(x+y+z) x y z}=2$
$\Rightarrow \frac{1}{2} b c \sin A=2 \Rightarrow b c=\frac{4}{\sin A}$.
Thus, $(x+y)^{2}+2(y+z)^{2}+3(z+x)^{2}$
$=a^{2}+2 b^{2}+3 c^{2}=3 b^{2}+4 c^{2}-2 b c \cos A$
$\geqslant 2 \sqrt{12} b c-2 b c \cos A$
$=(2 \sqrt{12}-2 \cos A) \frac{4}{\sin A}$
$=8(\sqrt{12}-\cos A) \frac{1}{\sin A}$.
Let $f(A)=(\sqrt{12}-\cos A) \frac{1}{\sin A}$. Then,
$f^{2}(A)=\frac{12-2 \sqrt{12} \cos A+\cos ^{2} A}{1-\cos ^{2} A}$
$=\frac{13-2 \sqrt{12} \cos A}{1-\cos ^{2} A}-1$.
Let $t=13-2 \sqrt{12} \cos A \Rightarrow \cos A=\frac{13-t}{2 \sqrt{12}}$.
Then $f^{2}(A)=\frac{48}{-t-\frac{121}{t}+26}-1$
$$
\geqslant \frac{48}{26-2 \sqrt{121}}-1=11 \text {. }
$$

Since $f \geqslant 0$, we have $f \geqslant \sqrt{11}$. Therefore,
$$
(x+y)^{2}+2(y+z)^{2}+3(z+x)^{2} \geqslant 8 \sqrt{11} \text {, }
$$

with equality when $3 b^{2}=4 c^{2}, 13-2 \sqrt{12} \cos A=11, b c \sin A=4$, i.e.,
$$
a=\sqrt{\frac{20}{\sqrt{11}}}, b=2 \sqrt{\frac{4}{\sqrt{11}}}, c=\sqrt{\frac{12}{\sqrt{11}}}, \cos A=\frac{\sqrt{3}}{6}
$$
11. (20 points) Given positive real numbers $x, y, z$ satisfy
$$
(x+y+z) x y z=4 \text {. }
$$

Find the minimum value of $(x+y)^{2}+2(y+z)^{2}+3(z+x)^{2}$.