Find all real numbers $t$ such that there exist non-negative real numbers $x, y, z$ satisfying
$$
\left\{\begin{array}{l}
3 x^{2}+3 z x+z^{2}=1, \\
3 y^{2}+3 y z+z^{2}=4, \\
x^{2}-x y+y^{2}=t .
\end{array}\right.
$$
Three equations are sequentially labeled as equations (1), (2), (3).
(Method 1) (2)-(1), we get
$$
3\left(y^{2}-x^{2}\right)+3(y-z) z=3,
$$

which simplifies to
$$
(y-x)(y+x)+(y-x) z=1.
$$

If $y-x \leqslant 0$, then equation (4) does not hold, so $y-x>0$. Let $a=y+x, b=y-x$, then $a, b>0$, and from equation (4) we know
$$
\text { (2) }+ \text { (1), we get } \begin{array}{c}
z=\frac{1}{b}-a . \\
3\left(x^{2}+y^{2}\right)+3(x+y) z+2 z^{2}=5 .
\end{array}
$$
(5)
(6)
Substituting equation (5) into equation (6), we have
$$
3 \cdot \frac{a^{2}+b^{2}}{2}+3 a\left(\frac{1}{b}-a\right)+2\left(\frac{1}{b}-a\right)^{2}=5,
$$

which simplifies to
$$
\frac{a^{2}+3 b^{2}}{2}+\frac{2-a b}{b^{2}}=5.
$$
(7)
Also, from $t=x^{2}-x y+y^{2}=\frac{a^{2}+3 b^{2}}{4}$, we know from equation (7) that
$$
t=\frac{5-\frac{2-a b}{b^{2}}}{2}.
$$

Rearranging equation (7) and completing the square, we get
$$
(1-a b)^{2}+3\left(\frac{5}{3}-b^{2}\right)^{2}=\frac{16}{3}.
$$

Let $m=2-a b, n=b^{2}$, then
$$
\begin{array}{l}
t=\frac{5-\frac{m}{n}}{2} \\
(m-1)^{2}+3\left(\frac{5}{3}-n\right)^{2}=\frac{16}{3}
\end{array}
$$

Now we find the range of $m, n$.
From equation (4), we know $a b=y^{2}-x^{2} \leqslant 1$, so $m=2-a b \geqslant 1$. From equations (1) and (2), we can solve for
$$
x=\frac{-3 z+\sqrt{12-3 z^{2}}}{6}, \quad y=\frac{-3 z+\sqrt{48-3 z^{2}}}{6}.
$$

Thus,
$$
y-x=\frac{\sqrt{48-3 z^{2}}-\sqrt{12-3 z^{2}}}{6}=\frac{6}{\sqrt{48-3 z^{2}}+\sqrt{12-3 z^{2}}}.
$$

From equation (1), we know $z \leqslant 1$, so $0 \leqslant z \leqslant 1$. Therefore,
$$
\frac{6}{\sqrt{48}+\sqrt{12}} \leqslant y-x \leqslant \frac{6}{\sqrt{45}+\sqrt{9}} \Rightarrow \frac{1}{\sqrt{3}} \leqslant b \leqslant \frac{\sqrt{5}-1}{2},
$$

Thus,
$$
\frac{1}{3} \leqslant n=b^{2} \leqslant \frac{3-\sqrt{5}}{2}.
$$

From equation (8) and the above analysis, the trajectory of $(m, n)$ in the $m O n$ plane is an elliptical arc from point $A\left(1, \frac{1}{3}\right)$ to point $B\left(\frac{\sqrt{5}+1}{2}, \frac{3-\sqrt{5}}{2}\right)$ (as shown in Figure 5). Calculating the $n$-coordinate of the tangent line from $O$ to the ellipse, we find the two $n$-coordinates are $1$ and $\frac{3}{7}$. Since $\frac{1}{3}<\frac{3-\sqrt{5}}{2}<\frac{3}{7}$, the lines from $O$ to any point on the ellipse between $A$ and $B$ are not tangents to the ellipse. Thus, the minimum and maximum values of $\frac{m}{n}$ are $\frac{1}{\frac{1}{3}}$ and $\frac{\frac{\sqrt{5}+1}{2}}{\frac{3-\sqrt{5}}{2}}$, i.e., 3 and $2+\sqrt{5}$. Therefore, the range of $\frac{m}{n}$ is $[3,2+\sqrt{5}]$, and thus the range of $t=\frac{5-\frac{m}{n}}{2}$ is $\left[\frac{3-\sqrt{5}}{2}, 1\right]$. The required $t$ is all real numbers satisfying $\frac{3-\sqrt{5}}{2} \leqslant t \leqslant 1$.
(Method 2) If the original system of equations has non-negative real solutions, then take a point $O$ in the plane, and draw three rays through $O$ such that the angles between them are $\frac{5 \pi}{6}, \frac{5 \pi}{6}, \frac{\pi}{3}$. On these three rays, take points $A, B, C$ such that $O A=z, O B=\sqrt{3} x, O C=\sqrt{3} y$. From equation (1), we know
$$
\begin{aligned}
A B^{2} & =O A^{2}+O B^{2}-2 O A \cdot O B \cos \angle A O B \\
& =z^{2}+3 x^{2}-2 z \cdot \sqrt{3} x \cdot\left(-\frac{\sqrt{3}}{2}\right) \\
& =z^{2}+3 x^{2}+3 x z=1.
\end{aligned}
$$

Thus, $A B=1$. Similarly, from equation (2), we know $A C=2$. From equation (3), we know
$$
\begin{aligned}
B C^{2} & =O B^{2}+O C^{2}-2 O B \cdot O C \cos \angle B O C \\
& =(\sqrt{3} x)^{2}+(\sqrt{3} y)^{2}-2(\sqrt{3} x)(\sqrt{3} y) \cdot \frac{1}{2} \\
& =3\left(x^{2}+y^{2}-x y\right)=3 t.
\end{aligned}
$$

Thus, $B C=\sqrt{3 t}$.
Since $\angle B A C \leqslant \angle B O C=\frac{\pi}{3}, \angle A C B \leqslant \angle A O B=\frac{5 \pi}{6}, \angle A B C \leqslant \angle A O C=\frac{5 \pi}{6}$, we have
$$
\begin{array}{l}
\frac{1}{2}=\cos \frac{\pi}{3} \leqslant \cos \angle B A C=\frac{1+4-3 t}{2 \times 1 \times 2}=\frac{5-3 t}{4}, \\
-\frac{\sqrt{3}}{2}=\cos \frac{5 \pi}{6} \leqslant \cos \angle A B C=\frac{1+3 t-4}{2 \sqrt{3} t}=\frac{3 t-3}{2 \sqrt{3 t}}, \\
-\frac{\sqrt{3}}{2}=\cos \frac{5 \pi}{6} \leqslant \cos \angle A C B=\frac{4+3 t-1}{2 \times 2 \times \sqrt{3} t}=\frac{3 t+3}{4 \sqrt{3 t}}.
\end{array}
$$

Thus, $\frac{3-\sqrt{5}}{2} \leqslant t \leqslant 1$. Also, $1 \leqslant \sqrt{3 t} \leqslant 3$, so $\frac{1}{3} \leqslant t \leqslant 3$, and thus $\frac{3-\sqrt{5}}{2} \leqslant t \leqslant 1$.
On the other hand, when $\frac{3-\sqrt{5}}{2} \leqslant t \leqslant 1$, from the above process, we know there exists a $\triangle A B C$ satisfying $A B=1, A C=2, B C=\sqrt{3} t$, and $\angle B A C \leqslant \frac{\pi}{3}, \angle A B C \leqslant \frac{5 \pi}{6}, \angle A B C \leqslant \frac{5 \pi}{6}$ (as shown in Figure 6). Thus, the arc with $A C$ as the chord and $\frac{5 \pi}{6}$ as the angle intersects the arc with $A B$ as the chord and $\frac{5 \pi}{6}$ as the angle inside $\triangle A B C$ at point $O$. Therefore, the solution to the original system of equations is $(x, y, z)=\left(\frac{O B}{\sqrt{3}}, \frac{O C}{\sqrt{3}}, O A\right)$.
Thus, the required $t$ is all real numbers satisfying $\frac{3-\sqrt{5}}{2} \leqslant t \leqslant 1$.