Let $$\vec{a}, \vec{b}, \vec{c}$$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and $$(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = 168$$ then $$|\vec{a}| + |\vec{b}| + |\vec{c}|$$ is equal to

We have three coplanar concurrent vectors $$\vec{a}, \vec{b}, \vec{c}$$ such that the angle between any two of them is the same. Since they lie in a plane and radiate from a common point, the only configuration where all pairwise angles are equal is when each pair makes an angle of $$120°$$ with every other pair (as $$\frac{360°}{3} = 120°$$).

Let $$|\vec{a}| = a$$, $$|\vec{b}| = b$$, and $$|\vec{c}| = c$$. We are given that the product of magnitudes satisfies $$abc = 14$$.

Since all three vectors are coplanar, every cross product among them points in the same direction — perpendicular to the common plane. Let $$\hat{n}$$ be the unit normal to this plane. With the angle between each pair being $$120°$$ and $$\sin 120° = \frac{\sqrt{3}}{2}$$, we have $$\vec{a} \times \vec{b} = \frac{\sqrt{3}}{2}\,ab\,\hat{n}$$, $$\vec{b} \times \vec{c} = \frac{\sqrt{3}}{2}\,bc\,\hat{n}$$, and $$\vec{c} \times \vec{a} = \frac{\sqrt{3}}{2}\,ca\,\hat{n}$$. (All cross products point along $$\hat{n}$$ or $$-\hat{n}$$ depending on orientation, but the dot products of parallel vectors use the same sign consistently.)

Now we compute each dot product in the given expression. We have $$(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2}\,ab \cdot \frac{\sqrt{3}}{2}\,bc = \frac{3}{4}\,ab^2c$$. Similarly, $$(\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) = \frac{3}{4}\,abc^2$$, and $$(\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = \frac{3}{4}\,a^2bc$$.

Adding all three terms gives $$\frac{3}{4}\,abc(a + b + c) = \frac{3}{4} \times 14 \times (a + b + c) = \frac{21}{2}(a + b + c)$$.

Setting this equal to 168, we obtain $$\frac{21}{2}(a + b + c) = 168$$, which yields $$a + b + c = \frac{168 \times 2}{21} = 16$$.

Hence, the correct answer is Option C.