The distance of the point $$(1, -2, 4)$$ from the plane passing through the point $$(1, 2, 2)$$ and perpendicular to the planes $$x - y + 2z = 3$$ and $$2x - 2y + z + 12 = 0$$, is:

To find the distance of the point $$(1, -2, 4)$$ from a plane passing through $$(1, 2, 2)$$ and perpendicular to the planes $$x - y + 2z = 3$$ and $$2x - 2y + z + 12 = 0$$, we first need the equation of the plane. Since the plane is perpendicular to two given planes, its normal vector must be perpendicular to the normal vectors of both given planes.

The normal vector of the first plane $$x - y + 2z - 3 = 0$$ is $$\vec{n_1} = (1, -1, 2)$$. The normal vector of the second plane $$2x - 2y + z + 12 = 0$$ is $$\vec{n_2} = (2, -2, 1)$$. The normal vector $$\vec{n}$$ of the required plane is the cross product of $$\vec{n_1}$$ and $$\vec{n_2}$$:

$$\vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & -2 & 1 \end{vmatrix}$$

Expanding the determinant:

$$\hat{i} \left[ (-1)(1) - (2)(-2) \right] - \hat{j} \left[ (1)(1) - (2)(2) \right] + \hat{k} \left[ (1)(-2) - (-1)(2) \right]$$

Computing each component:

For $$\hat{i}$$: $$(-1)(1) - (2)(-2) = -1 - (-4) = -1 + 4 = 3$$

For $$\hat{j}$$: $$-\left[ (1)(1) - (2)(2) \right] = -\left[ 1 - 4 \right] = -[-3] = 3$$

For $$\hat{k}$$: $$(1)(-2) - (-1)(2) = -2 - (-2) = -2 + 2 = 0$$

So $$\vec{n} = (3, 3, 0)$$. We can simplify this by dividing by 3, giving $$\vec{n} = (1, 1, 0)$$.

The plane passes through the point $$(1, 2, 2)$$. Using the point-normal form, the equation is:

$$(1)(x - 1) + (1)(y - 2) + (0)(z - 2) = 0$$

Simplifying:

$$x - 1 + y - 2 = 0 \implies x + y - 3 = 0$$

Now, we find the distance from the point $$(1, -2, 4)$$ to the plane $$x + y - 3 = 0$$. The general formula for the distance from a point $$(x_0, y_0, z_0)$$ to a plane $$ax + by + cz + d = 0$$ is:

$$\text{Distance} = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$$

Rewriting the plane equation as $$x + y + 0 \cdot z - 3 = 0$$, we have $$a = 1$$, $$b = 1$$, $$c = 0$$, $$d = -3$$. Substituting the point $$(1, -2, 4)$$:

$$|1 \cdot 1 + 1 \cdot (-2) + 0 \cdot 4 - 3| = |1 - 2 - 3| = |-4| = 4$$

The denominator is $$\sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$$. Thus, the distance is:

$$\frac{4}{\sqrt{2}} = \frac{4 \sqrt{2}}{2} = 2\sqrt{2}$$

Comparing with the options, $$2\sqrt{2}$$ corresponds to Option C. Hence, the correct answer is Option C.