If $$\hat{x}$$, $$\hat{y}$$ and $$\hat{z}$$ are three unit vectors in threedimensional space, then the minimum value of $$|\hat{x} + \hat{y}|^2 + |\hat{y} + \hat{z}|^2 + |\hat{z} + \hat{x}|^2$$

We are given three unit vectors $$\hat{x}$$, $$\hat{y}$$, and $$\hat{z}$$ in three-dimensional space and need to find the minimum value of the expression $$|\hat{x} + \hat{y}|^2 + |\hat{y} + \hat{z}|^2 + |\hat{z} + \hat{x}|^2$$.

First, recall that for any vector $$\vec{a}$$, $$|\vec{a}|^2 = \vec{a} \cdot \vec{a}$$. Using this, expand each term in the expression.

Start with $$|\hat{x} + \hat{y}|^2$$:

$$ |\hat{x} + \hat{y}|^2 = (\hat{x} + \hat{y}) \cdot (\hat{x} + \hat{y}) = \hat{x} \cdot \hat{x} + 2 \hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{y} $$

Since $$\hat{x}$$ and $$\hat{y}$$ are unit vectors, $$\hat{x} \cdot \hat{x} = 1$$ and $$\hat{y} \cdot \hat{y} = 1$$, so:

$$ |\hat{x} + \hat{y}|^2 = 1 + 2 \hat{x} \cdot \hat{y} + 1 = 2 + 2 \hat{x} \cdot \hat{y} $$

Similarly, for $$|\hat{y} + \hat{z}|^2$$:

$$ |\hat{y} + \hat{z}|^2 = (\hat{y} + \hat{z}) \cdot (\hat{y} \cdot \hat{z}) = \hat{y} \cdot \hat{y} + 2 \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{z} = 1 + 2 \hat{y} \cdot \hat{z} + 1 = 2 + 2 \hat{y} \cdot \hat{z} $$

And for $$|\hat{z} + \hat{x}|^2$$:

$$ |\hat{z} + \hat{x}|^2 = (\hat{z} + \hat{x}) \cdot (\hat{z} + \hat{x}) = \hat{z} \cdot \hat{z} + 2 \hat{z} \cdot \hat{x} + \hat{x} \cdot \hat{x} = 1 + 2 \hat{z} \cdot \hat{x} + 1 = 2 + 2 \hat{z} \cdot \hat{x} $$

Now, sum these expressions:

$$ |\hat{x} + \hat{y}|^2 + |\hat{y} + \hat{z}|^2 + |\hat{z} + \hat{x}|^2 = (2 + 2 \hat{x} \cdot \hat{y}) + (2 + 2 \hat{y} \cdot \hat{z}) + (2 + 2 \hat{z} \cdot \hat{x}) = 6 + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) $$

Let $$s = \hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}$$. The expression becomes:

$$ 6 + 2s $$

To minimize the entire expression, minimize $$s$$. Since $$\hat{x}$$, $$\hat{y}$$, and $$\hat{z}$$ are unit vectors, the dot products equal the cosines of the angles between them:

$$ \hat{x} \cdot \hat{y} = \cos \theta_{xy}, \quad \hat{y} \cdot \hat{z} = \cos \theta_{yz}, \quad \hat{z} \cdot \hat{x} = \cos \theta_{zx} $$

So, $$s = \cos \theta_{xy} + \cos \theta_{yz} + \cos \theta_{zx}$$.

Consider the magnitude of the sum $$\hat{x} + \hat{y} + \hat{z}$$:

$$ |\hat{x} + \hat{y} + \hat{z}|^2 = (\hat{x} + \hat{y} + \hat{z}) \cdot (\hat{x} + \hat{y} + \hat{z}) = \hat{x} \cdot \hat{x} + \hat{y} \cdot \hat{y} + \hat{z} \cdot \hat{z} + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) = 3 + 2s $$

Since $$|\hat{x} + \hat{y} + \hat{z}|^2 \geq 0$$, we have:

$$ 3 + 2s \geq 0 \quad \Rightarrow \quad s \geq -\frac{3}{2} $$

The minimum value of $$s$$ is $$-\frac{3}{2}$$, achieved when $$|\hat{x} + \hat{y} + \hat{z}|^2 = 0$$, meaning $$\hat{x} + \hat{y} + \hat{z} = \vec{0}$$. This occurs when the vectors are coplanar and equally spaced at 120 degrees to each other.

For example, take:

$$ \hat{x} = (1, 0, 0), \quad \hat{y} = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right), \quad \hat{z} = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}, 0\right) $$

Verify these are unit vectors:

Magnitude of $$\hat{y}$$: $$\sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$$, similarly for $$\hat{z}$$.

Compute dot products:

$$ \hat{x} \cdot \hat{y} = (1)\left(-\frac{1}{2}\right) + (0)\left(\frac{\sqrt{3}}{2}\right) = -\frac{1}{2} $$

$$ \hat{y} \cdot \hat{z} = \left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right) = \frac{1}{4} - \frac{3}{4} = -\frac{1}{2} $$

$$ \hat{z} \cdot \hat{x} = \left(-\frac{1}{2}\right)(1) + \left(-\frac{\sqrt{3}}{2}\right)(0) = -\frac{1}{2} $$

So, $$s = -\frac{1}{2} + (-\frac{1}{2}) + (-\frac{1}{2}) = -\frac{3}{2}$$.

Substitute $$s = -\frac{3}{2}$$ into the expression:

$$ 6 + 2s = 6 + 2\left(-\frac{3}{2}\right) = 6 - 3 = 3 $$

Verify with the example vectors:

$$ \hat{x} + \hat{y} = \left(1 - \frac{1}{2}, 0 + \frac{\sqrt{3}}{2}, 0\right) = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right), \quad |\hat{x} + \hat{y}|^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1 $$

$$ \hat{y} + \hat{z} = \left(-\frac{1}{2} - \frac{1}{2}, \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}, 0\right) = (-1, 0, 0), \quad |\hat{y} + \hat{z}|^2 = (-1)^2 = 1 $$

$$ \hat{z} + \hat{x} = \left(-\frac{1}{2} + 1, -\frac{\sqrt{3}}{2} + 0, 0\right) = \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}, 0\right), \quad |\hat{z} + \hat{x}|^2 = \left(\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1 $$

Sum: $$1 + 1 + 1 = 3$$.

Since $$s \geq -\frac{3}{2}$$, the expression $$6 + 2s \geq 6 + 2(-\frac{3}{2}) = 3$$, and we achieved 3, it is the minimum.

Hence, the correct answer is Option B.