The volume of the tetrahedron with $\hat{i}-\lambda \hat{j}+\hat{k}$, $\lambda \hat{i}-\hat{j}-\hat{k}$ and $\hat{i}+\hat{j}+\lambda \hat{k}$ as coterminous edges is $2$. If $\lambda$ is an integer, then $|\lambda \hat{i}-3 \lambda \hat{j}+3 \hat{k}|=$

Three vectors $\hat{i}-\hat{k}$,$\lambda \hat{i}+\hat{j}+(1-\lambda) \hat{k}$,and $\mu \hat{i}+\lambda \hat{j}+(1+\lambda-\mu) \hat{k}$ represent the coterminous edges of a parallelepiped. The volume of the parallelepiped depends on:

The volume of the parallelepiped formed by the vectors $\hat{i} + m \hat{j} + \hat{k}$,$\hat{j} + m \hat{k}$,and $m \hat{i} + \hat{k}$ becomes minimum when $m$ is

Consider $\overrightarrow{r}, \overrightarrow{a}, \overrightarrow{b}$ and $\overrightarrow{c}$ are non-zero vectors such that $\overrightarrow{r} \cdot \overrightarrow{a}=0$,$|\overrightarrow{r} \times \overrightarrow{b}|=|\overrightarrow{r}||\overrightarrow{b}|$,and $|\overrightarrow{r} \times \overrightarrow{c}|=|\overrightarrow{r}||\overrightarrow{c}|$. Then,the scalar triple product $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$ is:

$(a+b) \cdot(b+c) \times(a+b+c)$ is equal to

If three non-zero vectors are $a = a_1 i + a_2 j + a_3 k,$ $b = b_1 i + b_2 j + b_3 k$ and $c = c_1 i + c_2 j + c_3 k.$ If $c$ is the unit vector perpendicular to the vectors $a$ and $b$ and the angle between $a$ and $b$ is $\frac{\pi}{6},$ then $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|^2$ is equal to