\(\vec{a}, \vec{b}, \vec{c}\) are 3 vectors, such that \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\), \(|\vec{a}| = 1\), \(|\vec{b}| = 2\), \(|\vec{c}| = 3\), then \(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}\) is equal to
Number of correct statements is $k$. Then $2k$ is: I) $A=\{1,2,3,4,5,6,7\}$, $B=\{T\subseteq A:\text{either }1\notin T\text{ or }2\in T\}$, $C=\{T\subseteq A:\text{sum of elements is prime}\}$. Number of elements in $B\cup C$ is 107. II) If sum of $n$ terms of any sequence is quadratic in $n$, then the sequence is AP. III) $x=(8\sqrt{3}+13)^{13}$ and $y=(7\sqrt{2}+9)^9$: $[x]+[y]$ is even. IV) $(x+(x^3-1)^{1/2})^5+(x-(x^3-1)^{1/2})^5$ is polynomial of degree 7.