The area of the quadrilateral $ABCD$ with vertices $A(2, 1, 1)$,$B(1, 2, 5)$,$C(-2, -3, 5)$,and $D(1, -6, -7)$ is equal to

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + \hat{k}$,$\overrightarrow{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$,$\overrightarrow{c} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\overrightarrow{d}$ be a vector such that $\overrightarrow{b} \times \overrightarrow{d} = \overrightarrow{c} \times \overrightarrow{d}$ and $\overrightarrow{a} \cdot \overrightarrow{d} = 4$. Then $|(\overrightarrow{a} \times \overrightarrow{d})|^2$ is equal to . . . . . . .

Let $\vec{a}=2 \hat{i}-3 \hat{j}-5 \hat{k}$ and $\vec{b}=3 \hat{i}+2 \hat{j}-5 \hat{k}$ be two vectors and $\vec{r}$ be a vector in the plane of $\vec{a}$ and $\vec{b}$. If $\vec{r}$ is orthogonal to the vector $5 \hat{i}-2 \hat{j}+3 \hat{k}$ and the magnitude of $\vec{r}$ is $\sqrt{94}$,then $|\vec{r} \cdot \vec{b}|=$

The unit vector perpendicular to each of the vectors $\bar{a}+\bar{b}$ and $\bar{a}-\bar{b}$,where $\bar{a}=\hat{i}+\hat{j}+\hat{k}$ and $\bar{b}=3 \hat{i}-2 \hat{j}+5 \hat{k}$ is

$a, b, c, d$ are coplanar vectors, then $(a \times b) \times (c \times d)$ is equal to

Let $\bar{a} = \hat{i} + \hat{j} - \hat{k}$ and $\bar{c} = 5\hat{i} - 3\hat{j} + 2\hat{k}$. If $\bar{b} \times \bar{c} = \bar{a}$,then find $|\bar{b}|$.