Let $$x_0$$ be the point of local maxima of $$f(x) = \vec{a} \cdot (\vec{b} \times \vec{c})$$, where $$\vec{a} = x\hat{i} - 2\hat{j} + 3\hat{k}$$, $$\vec{b} = -2\hat{i} + x\hat{j} - \hat{k}$$ and $$\vec{c} = 7\hat{i} - 2\hat{j} + x\hat{k}$$. Then the value of $$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$$ at $$x = x_0$$ is:

We have the vector-valued function $$f(x)=\vec a\cdot(\vec b\times\vec c)$$ where

$$\vec a=x\hat i-2\hat j+3\hat k,\qquad \vec b=-2\hat i+x\hat j-\hat k,\qquad \vec c=7\hat i-2\hat j+x\hat k.$$

First we evaluate the cross product $$\vec b\times\vec c$$. Using the formula for the determinant of a $$3\times3$$ matrix,

$$ \vec b\times\vec c= \begin{vmatrix} \hat i&\hat j&\hat k\\ -2&x&-1\\ 7&-2&x \end{vmatrix}. $$

Expanding this determinant:

$$ \vec b\times\vec c= \hat i\bigl(x\cdot x-(-1)(-2)\bigr) -\hat j\bigl((-2)\cdot x-(-1)\cdot7\bigr) +\hat k\bigl((-2)(-2)-x\cdot7\bigr). $$

Simplifying each component one by one:

$$ \hat i:\;x^2-2,\qquad \hat j:\;-\bigl(-2x+7\bigr)=2x-7,\qquad \hat k:\;4-7x. $$

So

$$\vec b\times\vec c=(\,x^2-2\,)\hat i+(\,2x-7\,)\hat j+(\,4-7x\,)\hat k.$$

Next we compute the dot product $$\vec a\cdot(\vec b\times\vec c)$$. Using the component-wise dot-product formula,

$$ \vec a\cdot(\vec b\times\vec c)= x\,(x^2-2)+(-2)\,(2x-7)+3\,(4-7x). $$

Multiplying out each term:

$$ x(x^2-2)=x^3-2x, \qquad (-2)(2x-7)=-4x+14, \qquad 3(4-7x)=12-21x. $$

Adding them together gives

$$ f(x)=x^3-2x-4x+14+12-21x =x^3-27x+26. $$

To find local extrema we differentiate. The derivative is

$$ f'(x)=\frac{d}{dx}(x^3-27x+26)=3x^2-27 =3(x^2-9)=3(x-3)(x+3). $$

Setting $$f'(x)=0$$ yields critical points $$x=3$$ and $$x=-3$$. We now apply the second-derivative test. The second derivative is

$$ f''(x)=\frac{d}{dx}\bigl(3x^2-27\bigr)=6x. $$

At $$x=3$$ we have $$f''(3)=18\gt 0$$ (concave up), so $$x=3$$ is a local minimum. At $$x=-3$$ we have $$f''(-3)=-18\lt 0$$ (concave down), so $$x=-3$$ is the point of local maximum. Thus $$x_0=-3.$$

Now we need the value of $$\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a$$ at this $$x_0$$. We evaluate each dot product symbolically first.

Using $$\vec a=(x,-2,3),\ \vec b=(-2,x,-1),\ \vec c=(7,-2,x):$$

$$ \vec a\cdot\vec b =x(-2)+(-2)(x)+3(-1) =-2x-2x-3 =-4x-3. $$

$$ \vec b\cdot\vec c =(-2)(7)+x(-2)+(-1)(x) =-14-2x-x =-14-3x. $$

$$ \vec c\cdot\vec a =(7)(x)+(-2)(-2)+x(3) =7x+4+3x =10x+4. $$

Adding these three results:

$$ \vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a =(-4x-3)+(-14-3x)+(10x+4) =(-4x-3x+10x)+(-3-14+4) =3x-13. $$

Finally, substitute $$x=x_0=-3$$ obtained earlier:

$$ 3(-3)-13=-9-13=-22. $$

Hence, the correct answer is Option D.