Determine continuous functions $f:\mathbb{R}\to \mathbb{R}$ such that $\left( {{a}^{2}}+ab+{{b}^{2}} \right)\int\limits_{a}^{b}{f\left( x \right)dx=3\int\limits_{a}^{b}{{{x}^{2}}f\left( x \right)dx,}}$ for every $a,b\in \mathbb{R}$ .
1. Let \( P(a, b) \) be the assertion of the given equation:
   \[
   \left( a^2 + ab + b^2 \right) \int_a^b f(x) \, dx = 3 \int_a^b x^2 f(x) \, dx
   \]
   for every \( a, b \in \mathbb{R} \).

2. Consider the special case \( P(0, t) \):
   \[
   \left( 0^2 + 0 \cdot t + t^2 \right) \int_0^t f(x) \, dx = 3 \int_0^t x^2 f(x) \, dx
   \]
   Simplifying, we get:
   \[
   t^2 \int_0^t f(x) \, dx = 3 \int_0^t x^2 f(x) \, dx
   \]

3. Since \( f \) is continuous, we can differentiate both sides with respect to \( t \). Using the Leibniz rule for differentiation under the integral sign, we get:
   \[
   \frac{d}{dt} \left( t^2 \int_0^t f(x) \, dx \right) = \frac{d}{dt} \left( 3 \int_0^t x^2 f(x) \, dx \right)
   \]

4. Differentiating the left-hand side:
   \[
   \frac{d}{dt} \left( t^2 \int_0^t f(x) \, dx \right) = 2t \int_0^t f(x) \, dx + t^2 f(t)
   \]

5. Differentiating the right-hand side:
   \[
   \frac{d}{dt} \left( 3 \int_0^t x^2 f(x) \, dx \right) = 3 t^2 f(t)
   \]

6. Equating the derivatives, we get:
   \[
   2t \int_0^t f(x) \, dx + t^2 f(t) = 3 t^2 f(t)
   \]

7. Simplifying, we obtain:
   \[
   2t \int_0^t f(x) \, dx = 2 t^2 f(t)
   \]
   \[
   \int_0^t f(x) \, dx = t f(t)
   \]

8. Differentiating both sides with respect to \( t \) again, we get:
   \[
   f(t) = f(t) + t f'(t)
   \]
   \[
   t f'(t) = 0
   \]

9. Since \( t \) is arbitrary, we conclude that:
   \[
   f'(t) = 0
   \]

10. Therefore, \( f(t) \) must be a constant function. Let \( f(x) = C \) where \( C \in \mathbb{R} \).

11. To verify, substitute \( f(x) = C \) back into the original equation:
    \[
    \left( a^2 + ab + b^2 \right) \int_a^b C \, dx = 3 \int_a^b x^2 C \, dx
    \]
    \[
    \left( a^2 + ab + b^2 \right) C (b - a) = 3 C \int_a^b x^2 \, dx
    \]
    \[
    \left( a^2 + ab + b^2 \right) (b - a) = 3 \left[ \frac{x^3}{3} \right]_a^b
    \]
    \[
    \left( a^2 + ab + b^2 \right) (b - a) = (b^3 - a^3)
    \]
    \[
    \left( a^2 + ab + b^2 \right) (b - a) = (b - a)(a^2 + ab + b^2)
    \]
    This confirms that \( f(x) = C \) is indeed a solution.

The final answer is \( \boxed{ f(x) = C } \) where \( C \in \mathbb{R} \).