Let $$\vec{a} = \hat{i} + \hat{j} + \hat{k}$$, $$\vec{b} = -\hat{i} - 8\hat{j} + 2\hat{k}$$ and $$\vec{c} = 4\hat{i} + c_2\hat{j} + c_3\hat{k}$$ be three vectors such that $$\vec{b} \times \vec{a} = \vec{c} \times \vec{a}$$. If the angle between the vector $$\vec{c}$$ and the vector $$3\hat{i} + 4\hat{j} + \hat{k}$$ is $$\theta$$, then the greatest integer less than or equal to $$\tan^2 \theta$$ is:

Given $$\vec{a} = \hat{i} + \hat{j} + \hat{k}$$, $$\vec{b} = -\hat{i} - 8\hat{j} + 2\hat{k}$$, $$\vec{c} = 4\hat{i} + c_2\hat{j} + c_3\hat{k}$$.

$$\vec{b} \times \vec{a} = \vec{c} \times \vec{a}$$ implies $$(\vec{b} - \vec{c}) \times \vec{a} = \vec{0}$$.

This means $$\vec{b} - \vec{c}$$ is parallel to $$\vec{a}$$, so $$\vec{b} - \vec{c} = \lambda \vec{a}$$ for some scalar $$\lambda$$.

$$\vec{b} - \vec{c} = (-1 - 4, -8 - c_2, 2 - c_3) = (-5, -8 - c_2, 2 - c_3)$$

This must be parallel to $$(1, 1, 1)$$:

$$-5 = \lambda$$, $$-8 - c_2 = \lambda = -5$$, $$2 - c_3 = \lambda = -5$$

So $$c_2 = -3$$ and $$c_3 = 7$$.

$$\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$$

Now find the angle $$\theta$$ between $$\vec{c} = (4, -3, 7)$$ and $$\vec{d} = (3, 4, 1)$$:

$$\cos\theta = \frac{\vec{c} \cdot \vec{d}}{|\vec{c}||\vec{d}|} = \frac{12 - 12 + 7}{\sqrt{16 + 9 + 49}\sqrt{9 + 16 + 1}} = \frac{7}{\sqrt{74}\sqrt{26}}$$

$$\cos^2\theta = \frac{49}{74 \times 26} = \frac{49}{1924}$$

$$\tan^2\theta = \frac{1 - \cos^2\theta}{\cos^2\theta} = \frac{1924 - 49}{49} = \frac{1875}{49} \approx 38.265$$

$$\lfloor \tan^2\theta \rfloor = 38$$

The answer is $$\boxed{38}$$.