1. **Problem statement:** Given an equilateral triangle ABC with side length $2\sqrt{a}$, a point P inside the triangle satisfies $PA=2$, $PB=2\sqrt{3}$, and $PC=4$. We need to find the value of $a$. 2. **Key formula:** For any point P inside an equilateral triangle ABC with side length $s$, the following relation holds: $$PA^2 + PB^2 + PC^2 = s^2 + 3PG^2$$ where G is the centroid of the triangle and $PG$ is the distance from P to G. 3. Since the triangle is equilateral, the centroid G is also the center of the triangle, and the distance $PG$ is unknown but non-negative. 4. Substitute the given lengths: $$PA^2 = 2^2 = 4$$ $$PB^2 = (2\sqrt{3})^2 = 4 \times 3 = 12$$ $$PC^2 = 4^2 = 16$$ Sum: $$4 + 12 + 16 = 32$$ 5. Let the side length be $s = 2\sqrt{a}$, so $$s^2 = (2\sqrt{a})^2 = 4a$$ 6. Using the formula: $$PA^2 + PB^2 + PC^2 = s^2 + 3PG^2$$ $$32 = 4a + 3PG^2$$ 7. Since $PG^2 \geq 0$, the minimum value of $4a$ is when $PG=0$, i.e., when P coincides with the centroid G. 8. If P is the centroid, then: $$32 = 4a + 0 \implies 4a = 32 \implies a = 8$$ 9. Therefore, the value of $a$ is $8$. **Final answer:** $$\boxed{8}$$