Three. (20 points) The system of equations about $x, y, z$
$$
\left\{\begin{array}{l}
3 x+2 y+z=a, \\
x y+2 y z+3 z x=6
\end{array}\right.
$$

has real solutions $(x, y, z)$. Find the minimum value of the positive real number $a$.
From the first equation, we get $3 x+2 y=a-z$, and from the second equation, we get
$$
x y=6-z(3 x+2 y)=6-z(a-z)=z^{2}-a z+6 \text {. }
$$

From $(3 x+2 y)^{2} \geqslant 4 \times 3 x \times 2 y$, we get
$$
(a-z)^{2} \geqslant 24\left(z^{2}-a z+6\right) \text {, }
$$

which simplifies to $23 z^{2}-22 a z+144-a^{2} \leqslant 0$.
It is evident that the parabola opening upwards
$$
y=23 x^{2}-22 a x+144-a^{2}
$$

passes through the point $\left(z, 23 z^{2}-22 a z+144-a^{2}\right)$, which is not above the x-axis. Therefore, the parabola intersects the x-axis.
Thus, $\frac{\Delta}{4}=(11 a)^{2}-23\left(144-a^{2}\right) \geqslant 0$, which simplifies to
$a^{2} \geqslant 23, a \geqslant \sqrt{23}$ (since $a>0$).
When $a=\sqrt{23}$, we can take $x=\frac{2}{\sqrt{23}}, y=\frac{3}{\sqrt{23}}, z=\frac{11}{\sqrt{23}}$,

Therefore, the minimum value of $a$ is $\sqrt{23}$.