Find all triplets of real numbers $(x,y,z)$ that are solutions to the system of equations
$x^2+y^2+25z^2=6xz+8yz$
$ 3x^2+2y^2+z^2=240$
To find all triplets of real numbers \((x, y, z)\) that satisfy the given system of equations, we will solve the equations step by step.

1. **Rewrite the first equation:**
   \[
   x^2 + y^2 + 25z^2 = 6xz + 8yz
   \]
   Rearrange the terms to bring all terms to one side:
   \[
   x^2 + y^2 + 25z^2 - 6xz - 8yz = 0
   \]

2. **Complete the square for the terms involving \(x\) and \(z\):**
   \[
   x^2 - 6xz + 9z^2 + y^2 - 8yz + 16z^2 = 0
   \]
   Notice that:
   \[
   x^2 - 6xz + 9z^2 = (x - 3z)^2
   \]
   and
   \[
   y^2 - 8yz + 16z^2 = (y - 4z)^2
   \]
   Therefore, the equation becomes:
   \[
   (x - 3z)^2 + (y - 4z)^2 = 0
   \]

3. **Solve the equation \((x - 3z)^2 + (y - 4z)^2 = 0\):**
   Since the sum of two squares is zero, each square must be zero:
   \[
   (x - 3z)^2 = 0 \quad \text{and} \quad (y - 4z)^2 = 0
   \]
   This implies:
   \[
   x - 3z = 0 \quad \text{and} \quad y - 4z = 0
   \]
   Solving these, we get:
   \[
   x = 3z \quad \text{and} \quad y = 4z
   \]

4. **Substitute \(x = 3z\) and \(y = 4z\) into the second equation:**
   The second equation is:
   \[
   3x^2 + 2y^2 + z^2 = 240
   \]
   Substitute \(x = 3z\) and \(y = 4z\):
   \[
   3(3z)^2 + 2(4z)^2 + z^2 = 240
   \]
   Simplify:
   \[
   3 \cdot 9z^2 + 2 \cdot 16z^2 + z^2 = 240
   \]
   \[
   27z^2 + 32z^2 + z^2 = 240
   \]
   \[
   60z^2 = 240
   \]
   \[
   z^2 = 4
   \]
   \[
   z = 2 \quad \text{or} \quad z = -2
   \]

5. **Find the corresponding \(x\) and \(y\) values:**
   For \(z = 2\):
   \[
   x = 3z = 3 \cdot 2 = 6
   \]
   \[
   y = 4z = 4 \cdot 2 = 8
   \]
   Thus, one solution is \((x, y, z) = (6, 8, 2)\).

   For \(z = -2\):
   \[
   x = 3z = 3 \cdot (-2) = -6
   \]
   \[
   y = 4z = 4 \cdot (-2) = -8
   \]
   Thus, another solution is \((x, y, z) = (-6, -8, -2)\).

The final answer is \(\boxed{(6, 8, 2) \text{ or } (-6, -8, -2)}\).