If $a$ and $b$ are two unit vectors such that $a+b$ is also a unit vector,then $|a-b|^2=$

For three unit vectors $\vec{a}, \vec{b}, \vec{c}$ satisfying $|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$ and $|2\vec{a}+k\vec{b}+k\vec{c}|=3$, the positive value of $k$ is:

If the components of $\vec{a}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ along and perpendicular to $\vec{b}=3 \hat{i}+\hat{j}-\hat{k}$ respectively,are $\frac{16}{11}(3 \hat{i}+\hat{j}-\hat{k})$ and $\frac{1}{11}(-4 \hat{i}-5 \hat{j}-17 \hat{k})$,then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Let the position vectors of the points $P, Q, R$ and $S$ be $\vec{a}=\hat{i}+2 \hat{j}-5 \hat{k}$,$\vec{b}=3 \hat{i}+6 \hat{j}+3 \hat{k}$,$\vec{c}=\frac{17}{5} \hat{i}+\frac{16}{5} \hat{j}+7 \hat{k}$ and $\vec{d}=2 \hat{i}+\hat{j}+\hat{k}$,respectively. Then which of the following statements is true?

Suppose $\vec{a}+\vec{b}+\vec{c}=0$,$|\vec{a}|=3$,$|\vec{b}|=5$,$|\vec{c}|=7$. Then the angle between $\vec{a}$ and $\vec{b}$ is:

Let $\vec{a}$ and $\vec{b}$ be two vectors. Let $|\vec{a}|=1, |\vec{b}|=4$ and $\vec{a} \cdot \vec{b}=2$. If $\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}$,then the value of $\vec{b} \cdot \vec{c}$ is