Determine the sum of all possible surface areas of a cube two of whose vertices are $$(1, 2, 0)$$ and $$(3, 3, 2)$$.

Let the two given vertices of the cube be $$P(1,2,0)$$ and $$Q(3,3,2)$$.
Their position-vector difference is

$$\overrightarrow{PQ}= \langle 3-1,\; 3-2,\; 2-0\rangle = \langle 2,1,2\rangle.$$

Hence the distance between the two vertices is

$$|\overrightarrow{PQ}| = \sqrt{2^{2}+1^{2}+2^{2}} = \sqrt{9}=3.$$

In a cube of edge length $$s$$, three distinct inter-vertex distances occur:

• edge length   $$=s$$
• face diagonal length   $$=s\sqrt{2}$$
• body diagonal length   $$=s\sqrt{3}$$

Thus the measured distance $$3$$ can match any one of the three possibilities:

For each case we can indeed construct a cube:

• For Case 1, take $$\overrightarrow{PQ}$$ as one edge; choose any vector of length 3 perpendicular to $$\overrightarrow{PQ}$$ as the second edge, and their cross product (scaled) as the third.
• For Case 2, write $$\overrightarrow{PQ}= \mathbf{u}+\mathbf{v}$$ with $$\mathbf{u}\perp\mathbf{v}$$ and $$|\mathbf{u}|=|\mathbf{v}|=s_2$$; these give the two edges of one face.
• For Case 3, simply regard $$\overrightarrow{PQ}$$ as the body diagonal.

Therefore all three edge lengths are possible, and the corresponding surface areas are

$$A_1 = 6s_1^{2}=6(3)^{2}=54,$$
$$A_2 = 6s_2^{2}=6\left(\dfrac{3}{\sqrt{2}}\right)^{2}=6\cdot\dfrac{9}{2}=27,$$
$$A_3 = 6s_3^{2}=6(\sqrt{3})^{2}=6\cdot3=18.$$

The sum of all possible surface areas is

$$A_1+A_2+A_3 = 54+27+18 = 99.$$

Hence the required sum is 99.