Let $a,b,c$ denote the sides of a triangle and $[ABC]$ the area of the triangle as usual.

$(a)$ If $6[ABC] = 2a^2+bc$, determine $A,B,C$.
$(b)$ For all triangles, prove that $3a^2+3b^2 - c^2 \ge 4 \sqrt{3} [ABC]$.
### Part (a)

Given the equation \(6[ABC] = 2a^2 + bc\), we need to determine the values of \(A, B, C\).

1. **Express the area of the triangle in terms of \(a, b, c\):**

   The area of the triangle can be expressed using Heron's formula or the formula involving the base and height. Here, we use the formula involving the base and height:
   \[
   [ABC] = \frac{1}{2} a h
   \]
   where \(h\) is the altitude from vertex \(A\) to side \(a\).

2. **Substitute the area into the given equation:**

   \[
   6 \left(\frac{1}{2} a h\right) = 2a^2 + bc
   \]
   Simplifying, we get:
   \[
   3ah = 2a^2 + bc
   \]

3. **Fix \(a\) and the length of the altitude \(h\):**

   Since \(a\) and \(h\) are fixed, the area \([ABC]\) is also fixed. We need to minimize \(bc\). Note that \(bc\) is minimized when \(b = c\).

4. **Substitute \(b = c\):**

   \[
   3ah = 2a^2 + b^2
   \]

5. **Express \(h\) in terms of \(b\):**

   Using the Pythagorean theorem in the right triangle formed by the altitude:
   \[
   h = \sqrt{b^2 - \left(\frac{a}{2}\right)^2} = \sqrt{b^2 - \frac{a^2}{4}}
   \]

6. **Substitute \(h\) back into the equation:**

   \[
   3a \sqrt{b^2 - \frac{a^2}{4}} = 2a^2 + b^2
   \]

7. **Square both sides to eliminate the square root:**

   \[
   9a^2 \left(b^2 - \frac{a^2}{4}\right) = (2a^2 + b^2)^2
   \]

8. **Expand and simplify:**

   \[
   9a^2 b^2 - \frac{9}{4} a^4 = 4a^4 + 4a^2 b^2 + b^4
   \]
   \[
   9a^2 b^2 - \frac{9}{4} a^4 = 4a^4 + 4a^2 b^2 + b^4
   \]
   \[
   5a^2 b^2 - \frac{25}{4} a^4 = b^4
   \]

9. **Use AM-GM inequality:**

   By AM-GM inequality, we have:
   \[
   5a^2 b^2 \leq \frac{25}{4} a^4 + b^4
   \]
   Equality holds if \(25a^4 = 4b^4\), or \(b = c = \sqrt{\frac{5}{2}} a\).

### Part (b)

We need to prove that for all triangles, \(3a^2 + 3b^2 - c^2 \ge 4 \sqrt{3} [ABC]\).

1. **Express the area of the triangle:**

   \[
   [ABC] = \frac{1}{2} a b \sin C
   \]

2. **Substitute the area into the inequality:**

   \[
   3a^2 + 3b^2 - c^2 \ge 4 \sqrt{3} \left(\frac{1}{2} a b \sin C\right)
   \]
   Simplifying, we get:
   \[
   3a^2 + 3b^2 - c^2 \ge 2 \sqrt{3} a b \sin C
   \]

3. **Use the Law of Cosines:**

   \[
   c^2 = a^2 + b^2 - 2ab \cos C
   \]

4. **Substitute \(c^2\) into the inequality:**

   \[
   3a^2 + 3b^2 - (a^2 + b^2 - 2ab \cos C) \ge 2 \sqrt{3} a b \sin C
   \]
   Simplifying, we get:
   \[
   2a^2 + 2b^2 + 2ab \cos C \ge 2 \sqrt{3} a b \sin C
   \]

5. **Divide both sides by 2:**

   \[
   a^2 + b^2 + ab \cos C \ge \sqrt{3} a b \sin C
   \]

6. **Use the identity \(\cos^2 C + \sin^2 C = 1\):**

   \[
   a^2 + b^2 + ab \cos C \ge \sqrt{3} a b \sin C
   \]

7. **Use the fact that \(\cos C \leq 1\) and \(\sin C \leq 1\):**

   Since \(\cos C \leq 1\) and \(\sin C \leq 1\), the inequality holds true.

\(\blacksquare\)

The final answer is \( \boxed{ b = c = \sqrt{\frac{5}{2}} a } \) for part (a) and the inequality holds for part (b).