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 of $$(7, -3, -4)$$ from the plane containing $$(2, -3, 1)$$, $$(-1, 1, -2)$$, and $$(3, -4, 2)$$.

We start by finding two vectors in the plane, namely $$\vec{AB} = (-3, 4, -3)$$ and $$\vec{AC} = (1, -1, 1)$$, where $$\vec{AB} = (-1-2, 1+3, -2-1)$$ and $$\vec{AC} = (3-2, -4+3, 2-1)$$.

Next, we determine a normal vector to the plane by computing the cross product $$\vec{n} = \vec{AB} \times \vec{AC}$$: $$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 4 & -3 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(-3+3) + \hat{k}(3-4) = (1, 0, -1).$$

Using the point $$(2, -3, 1)$$ and the normal vector $$(1, 0, -1)$$, the plane equation is $$1(x-2) + 0(y+3) - 1(z-1) = 0$$ which simplifies to $$x - 2 - z + 1 = 0$$ and hence $$x - z - 1 = 0.$$

Finally, the distance from $$(7, -3, -4)$$ to the plane $$x - z - 1 = 0$$ is given by $$d = \frac{|7 - (-4) - 1|}{\sqrt{1 + 0 + 1}} = \frac{|7 + 4 - 1|}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}.$$

Therefore, the required distance is $$5\sqrt{2}$$.