If $a=x \hat{i}+y \hat{j}+z \hat{k}$, then $(a \times \hat{i}) \cdot(\hat{i}+\hat{j})+(a \times \hat{j}) \cdot(\hat{j}+\hat{k})+(a \times \hat{k}) \cdot(\hat{k}+\hat{i})=$

Let $\vec{c}$ be a vector coplanar with the unit vectors $\vec{a}$ and $\vec{b}$, and let $\vec{d}$ be the unit vector perpendicular to $\vec{a}$, $\vec{b}$, and $\vec{c}$. If $[\vec{a} \vec{b} \vec{d}] \vec{c} - [\vec{a} \vec{b} \vec{c}] \vec{d} = \hat{i} + 2\hat{j} + 2\hat{k}$ and the angle between $\vec{a}$ and $\vec{b}$ is $30^{\circ}$, then $|\vec{c}| =$

$a \cdot (a \times b) = $

The sum of all values of $\alpha$,for which the points whose position vectors $\hat{i}-2 \hat{j}+3 \hat{k}$,$2 \hat{i}-3 \hat{j}+4 \hat{k}$,$(\alpha+1) \hat{i}+2 \hat{k}$ and $9 \hat{i}+(\alpha-8) \hat{j}+6 \hat{k}$ are coplanar,is equal to

The number of distinct real values of $\lambda$,for which the vectors $-\lambda^2 \hat{i}+\hat{j}+\hat{k}$,$\hat{i}-\lambda^2 \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}-\lambda^2 \hat{k}$ are coplanar,is

Let $a, b$ and $c$ be distinct positive numbers. If the vectors $a \hat{i} + a \hat{j} + c \hat{k}$,$\hat{i} + \hat{k}$ and $c \hat{i} + c \hat{j} + b \hat{k}$ are coplanar,then $c$ is equal to: