The sum of the intercepts on the coordinate axes of the plane passing through the point (-2, -2, 2) and containing the line joining the points (1, -1, 2) and (1, 1, 1) is:

We need to find the plane passing through the point $$(-2, -2, 2)$$ and containing the line joining $$A(1, -1, 2)$$ and $$B(1, 1, 1)$$.

The direction vector of line $$AB$$ is $$\vec{AB} = B - A = (0, 2, -1)$$.

The vector from $$A$$ to the point $$P(-2, -2, 2)$$ is $$\vec{AP} = (-3, -1, 0)$$.

The normal to the plane is perpendicular to both $$\vec{AB}$$ and $$\vec{AP}$$, so $$\vec{n} = \vec{AB} \times \vec{AP}$$.

Computing the cross product: $$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -1 \\ -3 & -1 & 0 \end{vmatrix}$$.

$$\vec{n} = \hat{i}(2 \cdot 0 - (-1)(-1)) - \hat{j}(0 \cdot 0 - (-1)(-3)) + \hat{k}(0 \cdot (-1) - 2 \cdot (-3))$$.

$$\vec{n} = \hat{i}(0 - 1) - \hat{j}(0 - 3) + \hat{k}(0 + 6) = (-1, 3, 6)$$.

The equation of the plane passing through $$A(1, -1, 2)$$ with normal $$(-1, 3, 6)$$ is: $$-1(x - 1) + 3(y + 1) + 6(z - 2) = 0$$.

Expanding: $$-x + 1 + 3y + 3 + 6z - 12 = 0$$, which simplifies to $$-x + 3y + 6z - 8 = 0$$, or $$x - 3y - 6z + 8 = 0$$.

To find the intercepts, set two variables to zero at a time. The x-intercept ($$y=0, z=0$$): $$x = -8$$. The y-intercept ($$x=0, z=0$$): $$y = \frac{8}{3}$$. The z-intercept ($$x=0, y=0$$): $$z = \frac{4}{3}$$.

The sum of the intercepts is $$-8 + \frac{8}{3} + \frac{4}{3} = -8 + \frac{12}{3} = -8 + 4 = -4$$.

The correct answer is Option D: $$-4$$.