For $$a\in\mathbb{R}$$, the solution set of $$x^2-3(a+1)x+a^2+a-1\le0$$ is an interval of length $$4$$. What is the sum of the squares of the different values of $$a$$?

The inequality is given by $$x^2 - 3(a + 1)x + a^2 + a - 1 \le 0$$.

Let the roots of the corresponding quadratic equation $$x^2 - 3(a + 1)x + a^2 + a - 1 = 0$$ be $$x_1$$ and $$x_2$$.

The solution set for the quadratic expression being less than or equal to zero is the closed interval $$[x_1, x_2]$$.

The length of this interval is given as $$4$$, which means:

$$\vert{}x_2 - x_1\vert{} = 4$$

Using the relation for the difference of roots of a quadratic equation, $$\vert{}x_2 - x_1\vert{} = \frac{\sqrt{D}}{\vert{}A\vert{}}$$, where $$D$$ is the discriminant:

$$\sqrt{D} = 4 \implies D = 16$$

Let us calculate the discriminant $$D$$ of the quadratic expression:

$$D = [-3(a + 1)]^2 - 4(1)(a^2 + a - 1)$$

$$D = 9(a^2 + 2a + 1) - 4a^2 - 4a + 4$$

$$D = 9a^2 + 18a + 9 - 4a^2 - 4a + 4$$

$$D = 5a^2 + 14a + 13$$

Equating the discriminant to $$16$$:

$$5a^2 + 14a + 13 = 16$$

$$5a^2 + 14a - 3 = 0$$

Solving this quadratic equation for $$a$$ by factoring:

$$5a^2 + 15a - a - 3 = 0$$

$$5a(a + 3) - 1(a + 3) = 0$$

$$(5a - 1)(a + 3) = 0$$

This gives two possible values for $$a$$:

$$a_1 = \frac{1}{5}, \quad a_2 = -3$$

We need to find the sum of the squares of these different values of $$a$$:

$$a_1^2 + a_2^2 = \left(\frac{1}{5}\right)^2 + (-3)^2 = \frac{1}{25} + 9 = \frac{1 + 225}{25} = \frac{226}{25}$$

The correct option is C.