To solve the problem, we need to determine the vectors normal to the planes containing given vectors and find the angle between those normal vectors.

First, let's find the vectors normal to the plane containing vectors $$\overrightarrow{A}$$ and $$\overrightarrow{B}$$. The normal vector can be calculated using the cross product:

$$\overrightarrow{A} \times \overrightarrow{B} = (\widehat{i} + \widehat{j}) \times (\widehat{j} + \widehat{k})$$.

Using the properties of the cross product:

$$\begin{aligned} \overrightarrow{A} \times \overrightarrow{B} &= (\widehat{i} \times \widehat{j}) + (\widehat{i} \times \widehat{k}) + (\widehat{j} \times \widehat{j}) + (\widehat{j} \times \widehat{k}) \\ &= \widehat{k} - \widehat{j} + 0 + \widehat{i} \\ &= \widehat{i} - \widehat{j} + \widehat{k}. \end{aligned}$$

So, the vector normal to the plane containing vectors $$\overrightarrow{A}$$ and $$\overrightarrow{B}$$ is $$\overrightarrow{N}_{1} = \widehat{i} - \widehat{j} + \widehat{k}$$.

Next, let's find the vector normal to the plane containing vectors $$\overrightarrow{A}$$ and $$\overrightarrow{C}$$:

$$\overrightarrow{A} \times \overrightarrow{C} = (\widehat{i} + \widehat{j}) \times (-\widehat{i} + \widehat{j})$$.

Using the properties of the cross product:

$$\begin{aligned} \overrightarrow{A} \times \overrightarrow{C} &= (\widehat{i} \times -\widehat{i}) + (\widehat{i} \times \widehat{j}) + (\widehat{j} \times -\widehat{i}) + (\widehat{j} \times \widehat{j}) \\ &= 0 + \widehat{k} - \widehat{k} + 0 \\ &= \widehat{k} - \widehat{k} + 0 \\ &= 2 \widehat{k}. \end{aligned}$$

So, the vector normal to the plane containing vectors $$\overrightarrow{A}$$ and $$\overrightarrow{C}$$ is $$\overrightarrow{N}_{2} = 2 \widehat{k}$$. But we just need the direction of this vector, not its magnitude, so we can simplify it to $$\widehat{k}$$.

Now, to find the angle between the two normal vectors $$\overrightarrow{N}_{1} = \widehat{i} - \widehat{j} + \widehat{k}$$ and $$\overrightarrow{N}_{2} = \widehat{k}$$, we use the dot product formula:

$$\cos(\theta) = \frac{\overrightarrow{N}_{1} \cdot \overrightarrow{N}_{2}}{||\overrightarrow{N}_{1}|| ||\overrightarrow{N}_{2}||}$$

First, calculate the dot product:

$$\overrightarrow{N}_{1} \cdot \overrightarrow{N}_{2} = (\widehat{i} - \widehat{j} + \widehat{k}) \cdot \widehat{k} = 0 + 0 + 1 = 1$$

Next, find the magnitude of the vectors:

$$||\overrightarrow{N}_{1}|| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$$

$$||\overrightarrow{N}_{2}|| = \sqrt{0^2 + 0^2 + 1^2} = 1$$

Now substitute these into the cosine formula:

$$\cos(\theta) = \frac{1}{\sqrt{3} \times 1} = \frac{1}{\sqrt{3}} = \sqrt{\frac{1}{3}}$$

Therefore, $$\cos^{-1}(\sqrt{\frac{1}{3}}) = \cos^{-1}(\frac{1}{\sqrt{3}})$$ indicates $$x = 3$$.

So, the value of $$x$$ is:

3