The distance of the point $$(7, -3, -4)$$ from the plane containing the points $$(2, -3, 1)$$, $$(-1, 1, -2)$$ and $$(3, -4, 2)$$ is equal to:

We need to find the distance from $$(7,-3,-4)$$ to the plane determined by the points $$(2,-3,1)$$, $$(-1,1,-2)$$ and $$(3,-4,2)$$. To begin with, we form two vectors in the plane: $$\vec{AB} = (-1-2, 1+3, -2-1) = (-3, 4, -3)$$ and $$\vec{AC} = (3-2, -4+3, 2-1) = (1, -1, 1)$$.

Next, the normal vector to the plane is found by taking the cross product of these vectors, which yields:
$$\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 4 & -3 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(-3+3) + \hat{k}(3-4) = (1, 0, -1)$$.

Since we know a point $$(2,-3,1)$$ on the plane and the normal vector $$(1,0,-1)$$, the equation of the plane can be written as $$1(x-2) + 0(y+3) - 1(z-1) = 0$$, which simplifies to $$x - z - 1 = 0$$.

Finally, the distance from the point $$(7,-3,-4)$$ to this plane is given by the formula:
$$d = \frac{|7 - (-4) - 1|}{\sqrt{1+0+1}} = \frac{|10|}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}$$. Therefore, the correct answer is Option C: $$5\sqrt{2}$$.