The value of $a$,so that the volume of the parallelepiped formed by $\hat{i} + a \hat{j} + \hat{k}$,$\hat{j} + a \hat{k}$,and $a \hat{i} + \hat{k}$ becomes minimum is

If the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+2\hat{k}$ and $\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$ are coplanar, then $x=$

If $\bar{a} = \hat{i} - \hat{j}$,$\bar{b} = \hat{j} - \hat{k}$,and $\bar{c} = \hat{k} - \hat{i}$,then a unit vector $\bar{d}$ such that $\bar{a} \cdot \bar{d} = 0$ and $[\bar{b} \bar{c} \bar{d}] = 0$ is:

$A$ unit vector $\vec{e} = a \hat{i} + b \hat{j} + c \hat{k}$ is coplanar with the vectors $\hat{i} - 3 \hat{j} + 5 \hat{k}$ and $3 \hat{i} + \hat{j} - 5 \hat{k}$. If $\vec{e}$ is perpendicular to the vector $\hat{i} + \hat{j} + \hat{k}$, then $2 a^2 + 3 b^2 + 4 c^2 =$

If the vectors $a \hat{i}+\hat{j}+\hat{k}$,$\hat{i}+b \hat{j}+\hat{k}$,and $\hat{i}+\hat{j}+c \hat{k}$ are coplanar,where $(a, b, c \neq 1)$,then the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$