To determine the values of $ a $ for which the quadratic expression $ x^2 - (3a - 1)x + 2a^2 + 2a - 11 $ is always positive, we need to analyze the quadratic form and ensure that its discriminant is negative. This will ensure that the quadratic has no real roots and hence does not change sign, provided that the coefficient of $ x^2 $ is positive (which it is, being 1).

In general, a quadratic equation of the form $ Ax^2 + Bx + C $ is always positive if its discriminant ($ \Delta $) is less than zero. The discriminant for a quadratic equation $ Ax^2 + Bx + C $ is given by:

$$\Delta = B^2 - 4AC$$

For our quadratic expression $ x^2 - (3a - 1)x + 2a^2 + 2a - 11 $, we identify:

$ A = 1 $
$ B = -(3a - 1) $
$ C = 2a^2 + 2a - 11 $

Substituting $ A $, $ B $, and $ C $ into the discriminant formula, we get:

$$\Delta = (-(3a - 1))^2 - 4(1)(2a^2 + 2a - 11)$$

Simplifying $ \Delta $, we have:

$$\Delta = (3a - 1)^2 - 4(2a^2 + 2a - 11)$$

Expanding these terms:

$$\Delta = (3a - 1)(3a - 1) - 4(2a^2 + 2a - 11)$$

$$\Delta = 9a^2 - 6a + 1 - (8a^2 + 8a - 44)$$

$$\Delta = 9a^2 - 6a + 1 - 8a^2 - 8a + 44$$

$$\Delta = a^2 - 14a + 45$$

Now, we need to find when this discriminant is less than zero:

$$a^2 - 14a + 45 < 0$$

To solve this inequality, we first find the roots of the corresponding equation:

$$a^2 - 14a + 45 = 0$$

Using the quadratic formula $ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ with $ a = 1 $, $ b = -14 $, and $ c = 45 $:

$$a = \frac{14 \pm \sqrt{196 - 180}}{2}$$

$$a = \frac{14 \pm \sqrt{16}}{2}$$

$$a = \frac{14 \pm 4}{2}$$

This gives us two roots:

$$a = \frac{18}{2} = 9$$
$$a = \frac{10}{2} = 5$$

Thus, the discriminant is less than zero for values of $ a $ that lie between these roots:

$$5 < a < 9$$

Therefore, the quadratic expression $ x^2 - (3a - 1)x + 2a^2 + 2a - 11 $ is always positive for:

$$5 < a < 9$$