Let $$\vec{a}$$ and $$\vec{b}$$ be two unit vectors such that $$|\vec{a} + \vec{b}| = \sqrt{3}$$. If $$\vec{c} = \vec{a} + 2\vec{b} + (\vec{a} \times \vec{b})$$, then $$2|\vec{c}|$$ is equal to:

Since $$\vec{a}$$ and $$\vec{b}$$ are unit vectors, $$|\vec{a}| = |\vec{b}| = 1$$. Expanding $$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2\vec{a}\cdot\vec{b} + |\vec{b}|^2 = 2 + 2\vec{a}\cdot\vec{b} = 3$$, which gives $$\vec{a}\cdot\vec{b} = \dfrac{1}{2}$$.

Now, $$\vec{c} = \vec{a} + 2\vec{b} + 3(\vec{a} \times \vec{b})$$. Since the cross product $$\vec{a} \times \vec{b}$$ is perpendicular to both $$\vec{a}$$ and $$\vec{b}$$, it is also perpendicular to every linear combination of $$\vec{a}$$ and $$\vec{b}$$. In particular, $$(\vec{a} + 2\vec{b}) \cdot (\vec{a} \times \vec{b}) = 0$$. Therefore $$|\vec{c}|^2 = |\vec{a} + 2\vec{b}|^2 + 9|\vec{a} \times \vec{b}|^2$$.

Computing the first term: $$|\vec{a} + 2\vec{b}|^2 = |\vec{a}|^2 + 4\vec{a}\cdot\vec{b} + 4|\vec{b}|^2 = 1 + 4 \cdot \dfrac{1}{2} + 4 = 7$$.

For the cross product magnitude: $$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2 = 1 - \dfrac{1}{4} = \dfrac{3}{4}$$.

So $$|\vec{c}|^2 = 7 + 9 \cdot \dfrac{3}{4} = 7 + \dfrac{27}{4} = \dfrac{55}{4}$$, giving $$|\vec{c}| = \dfrac{\sqrt{55}}{2}$$. Therefore $$2|\vec{c}| = \sqrt{55}$$.