Given that $$\vec{a} = \vec{b} \times \vec{c}$$, we can find the magnitudes of $$\vec{a}$$ and $$\vec{c}$$:

$$|\vec{a}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11}$$

$$|\vec{c}| = \sqrt{2^2 + (-3)^2 + 3^2} = \sqrt{22}$$

We know that the magnitude of the cross product of two vectors is equal to the product of the magnitudes of the vectors and the sine of the angle between them :

$$|\vec{a}| = |\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin\theta$$

Plugging in the known values :

$$\sqrt{11} = \sqrt{50}\sqrt{22}\sin\theta$$

Solving for the sine of the angle between the vectors :

$$\sin\theta = \frac{1}{10}$$

Now we can find $$|\vec{b} + \vec{c}|^2$$ using the formula :

$$|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c}$$

We have the dot product $$\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\theta$$, and we can use the relationship between sine and cosine: $$\cos\theta = \sqrt{1 - \sin^2\theta} = \frac{\sqrt{99}}{10}$$.

Substitute the values into the formula :

$$|\vec{b} + \vec{c}|^2 = 50 + 22 + 2\sqrt{50}\sqrt{22}\frac{\sqrt{99}}{10} = 72 + 66$$

Finally, we need to find the absolute value of the difference :

$$|72 - |\vec{b} + \vec{c}|^2| = |72 - (72 + 66)| = 66$$