$\vec{a}, \vec{b}, \vec{c}$ are three unit vectors such that $|\vec{a}+\vec{b}+\vec{c}|=1$ and $\vec{a}$ is perpendicular to $\vec{b}$. If $\vec{c}$ makes angles $\alpha, \beta$ with $\vec{a}, \vec{b}$ respectively, then $\cos \alpha+\cos \beta=$

If $\bar{a}, \bar{b}, \bar{c}$ are three vectors which are perpendicular to $\bar{b}+\bar{c}, \bar{c}+\bar{a}$,and $\bar{a}+\bar{b}$ respectively,such that $|\bar{a}|=2, |\bar{b}|=3, |\bar{c}|=4$,then $|\bar{a}+\bar{b}+\bar{c}|=$

Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=2 \hat{j}-3 \hat{k}$. If $\vec{b}=\vec{c}-\vec{d}$, $\vec{a}$ is parallel to $\vec{c}$, and $\vec{a}$ is perpendicular to $\vec{d}$, then $\vec{c}+\vec{d}=$

Let $a$,$b$,and $c$ be $3$ non-zero vectors such that no $2$ of these are collinear. If vector $a + 2b$ is collinear with $c$ and $b + 3c$ is collinear with $a$,then $a + 2b + 6c$ equals:

The vector $\vec{a} = \alpha \hat{i} + 2\hat{j} + \beta \hat{k}$ lies in the plane of $\vec{b} = \hat{i} + \hat{j}$ and $\vec{c} = \hat{j} + \hat{k}$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Find the possible values of $\alpha$ and $\beta$.

Let $\vec{a}=4 \hat{i}+3 \hat{j}$ and $\vec{b}=3 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\vec{c}$ is a vector such that $\vec{c} \cdot(\vec{a} \times \vec{b})+25=0, \vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=4$ and the projection of $\vec{c}$ on $\vec{a}$ is $1$. Then,the projection of $\vec{c}$ on $\vec{b}$ equals: