Find all real values $t$ for which there exist real numbers $x$, $y$, $z$ satisfying :
$3x^2 + 3xz + z^2 = 1$ ,
 $3y^2 + 3yz + z^2 = 4$,
 $x^2 - xy + y^2 = t$.
Given the equations:
\[ 3x^2 + 3xz + z^2 = 1 \quad\quad (1) \]
\[ 3y^2 + 3yz + z^2 = 4 \quad\quad (2) \]
\[ x^2 - xy + y^2 = t \quad\quad (3) \]

We need to find all real values of \( t \) for which there exist real numbers \( x \), \( y \), and \( z \) satisfying these equations.

1. **Multiply equations (1) and (2) by 4:**
   \[ 4(3x^2 + 3xz + z^2) = 4 \cdot 1 \]
   \[ 4(3y^2 + 3yz + z^2) = 4 \cdot 4 \]

   This gives:
   \[ 12x^2 + 12xz + 4z^2 = 4 \quad\quad (4) \]
   \[ 12y^2 + 12yz + 4z^2 = 16 \quad\quad (5) \]

2. **Add equations (4) and (5):**
   \[ 12x^2 + 12xz + 4z^2 + 12y^2 + 12yz + 4z^2 = 4 + 16 \]
   \[ 12x^2 + 12xz + 4z^2 + 12y^2 + 12yz + 4z^2 = 20 \]
   \[ 12x^2 + 12xz + 12y^2 + 12yz + 8z^2 = 20 \quad\quad (6) \]

3. **Subtract equation (3) multiplied by 2 from equation (6):**
   \[ 12x^2 + 12xz + 12y^2 + 12yz + 8z^2 - 2(x^2 - xy + y^2) = 20 - 2t \]
   \[ 12x^2 + 12xz + 12y^2 + 12yz + 8z^2 - 2x^2 + 2xy - 2y^2 = 20 - 2t \]
   \[ 10x^2 + 12xz + 10y^2 + 12yz + 8z^2 + 2xy = 20 - 2t \]

4. **Rearrange and factor the left-hand side:**
   \[ 10x^2 + 2xy + 10y^2 + 12xz + 12yz + 8z^2 = 20 - 2t \]
   \[ (3x + 2z)^2 + (3y + 2z)^2 + (x + y)^2 = 20 - 2t \]

5. **Since the left-hand side is a sum of squares, it must be non-negative:**
   \[ (3x + 2z)^2 + (3y + 2z)^2 + (x + y)^2 \geq 0 \]
   \[ 20 - 2t \geq 0 \]
   \[ 2t \leq 20 \]
   \[ t \leq 10 \]

Therefore, the real values of \( t \) for which there exist real numbers \( x \), \( y \), and \( z \) satisfying the given equations are all \( t \leq 10 \).

The final answer is \( \boxed{ t \leq 10 } \).