The area of the parallelogram for which the vectors $\hat{i}+\hat{j}+2 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$ are adjacent sides is equal to

Let $\bar{a}, \bar{b}, \bar{c}$ be vectors such that $\bar{a} \neq \bar{o}, \bar{b} \neq \bar{o}, \bar{a} \times \bar{c} = \bar{b}$ and $\bar{b} \times \bar{c} = \bar{a}$. Then:

$\vec{a}=\hat{i}+\hat{j}-2 \hat{k}$, $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}+\hat{j}-\hat{k}$ are three vectors. If $\vec{d}$ is a normal to the plane of $\vec{a}$ and $\vec{b}$ and $\vec{d} \cdot \vec{c}=2$, then $|\vec{d}|=$

The area of a parallelogram whose adjacent sides are given by the vectors $i + 2j + 3k$ and $-3i - 2j + k$ (in square units) is

If $a=2 \hat{i}+\hat{j}-3 \hat{k}$, $b=\hat{i}-2 \hat{j}+3 \hat{k}$, $c=-\hat{i}+\hat{j}-4 \hat{k}$ and $d=\hat{i}+\hat{j}+2 \hat{k}$, then $(a \times b) \times(c \times d)=$

If $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ are unit vectors such that $\bar{a} \cdot \bar{b} = \frac{1}{2}$,$\bar{c} \cdot \bar{d} = \frac{1}{2}$ and the angle between $\bar{a} \times \bar{b}$ and $\bar{c} \times \bar{d}$ is $\frac{\pi}{6}$,then the value of $|[\bar{a} \bar{b} \bar{d}] \bar{c} - [\bar{a} \bar{b} \bar{c}] \bar{d}| = $