Let $\vec{a}, \vec{b}, \vec{c}$ be vectors of lengths $3, 4, 5$ respectively. Let $\vec{a}$ be perpendicular to $\vec{b}+\vec{c}$,$\vec{b}$ be perpendicular to $\vec{c}+\vec{a}$,and $\vec{c}$ be perpendicular to $\vec{a}+\vec{b}$. Then the length of vector $\vec{a}+\vec{b}+\vec{c}$ is:

$\bar{a}, \bar{b}, \bar{c}$ are nonzero vectors such that $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$,$|\bar{a}|=1, |\bar{b}|=2, |\bar{c}|=1$ and $\bar{b} \cdot \bar{c}=1$. There is a nonzero vector $\bar{d}$ coplanar with $\bar{a}+\bar{b}$ and $2\bar{b}-\bar{c}$. If $\bar{d} \cdot \bar{a}=1$,then $|\bar{d}|^2=$ (Note that $x$ and $y$ are parameters involved when we write $\bar{d}=x(\bar{a}+\bar{b})+y(2\bar{b}-\bar{c})$)

If $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + \hat{k}$,$\vec{c} = \hat{i} + \hat{j} + \hat{k}$ and let $\vec{d}$ be such that $\vec{a} \times \vec{b} = \vec{d} \times \vec{b}$ and $\vec{d} \cdot \vec{c} = 8$,then the value of $\vec{d} \cdot \vec{b}$ is:

If the vectors $\vec{a} = \hat{i} - 2x\hat{j} - 3y\hat{k}$ and $\vec{b} = \hat{i} + 3x\hat{j} + 2y\hat{k}$ are perpendicular to each other,find the locus of the point $(x, y)$.

If $\bar{a}$ and $\bar{b}$ are not perpendicular to each other,$\bar{r} \times \bar{a} = \bar{b} \times \bar{a}$ and $\bar{r} \cdot \bar{c} = 0$,then $\bar{r} =$

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: