Example 2 The system of equations in $x, y, z$
$$
\left\{\begin{array}{l}
3 x+2 y+z=a, \\
x y+2 y z+3 z x=6
\end{array}\right.
$$

has real solutions $(x, y, z)$. Find the minimum value of the positive real number $a$.
Solving the system of equations, we get
$$
3 x+2 y=a-z, 3 x \times 2 y=6\left(z^{2}-a z+6\right) \text {. }
$$

It is easy to see that $3 x$ and $2 y$ are the two real roots of the quadratic equation
$$
t^{2}-(a-z) t+6\left(z^{2}-a z+6\right)=0
$$

Thus, $\Delta=[-(a-z)]^{2}-24\left(z^{2}-a z+6\right) \geqslant 0$, which simplifies to
$$
23 z^{2}-22 a z+144-a^{2} \leqslant 0 \text {. }
$$

Since $z$ is a real number, we have
$$
\Delta^{\prime}=(-22 a)^{2}-4 \times 23\left(144-a^{2}\right) \geqslant 0 \text {. }
$$

Solving this, we get $a \geqslant \sqrt{23}$.
Therefore, the minimum value of the positive real number $a$ is $\sqrt{23}$.