Let $A(2, 3, 5)$ and $C(-3, 4, -2)$ be opposite vertices of a parallelogram $ABCD$. If the diagonal $\overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$,then the area of the parallelogram is equal to:

The area of the triangle having vertices as $i - 2j + 3k,$ $- 2i + 3j - k,$ and $4i - 7j + 7k$ is

Let $ABC$ be a triangle of area $15 \sqrt{2}$ and the vectors $\overrightarrow{AB}=\hat{i}+2 \hat{j}-7 \hat{k}$,$\overrightarrow{BC}=a \hat{i}+b \hat{j}+c \hat{k}$ and $\overrightarrow{AC}=6 \hat{i}+d \hat{j}-2 \hat{k}$,where $d>0$. Then the square of the length of the largest side of the triangle $ABC$ is:

$\vec{u}, \vec{v}, \vec{w}$ are three unit vectors. Let $\vec{p}=\vec{u}+\vec{v}+\vec{w}$ and $\vec{q}=\vec{u} \times(\vec{v} \times \vec{w})$. If $\vec{p} \cdot \vec{u}=\frac{3}{2}, \vec{p} \cdot \vec{v}=\frac{7}{4}, |\vec{p}|=2$ and $\vec{v}=K \vec{q}$,then $K=$

If the diagonals of a parallelogram are represented by the vectors $3\hat{i} + \hat{j} - 2\hat{k}$ and $\hat{i} + 3\hat{j} - 4\hat{k}$,then its area in square units is:

If $\overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b}$,$\overrightarrow{v}=\overrightarrow{a}+\overrightarrow{b}$,$|\overrightarrow{a}|=|\overrightarrow{b}|=2$,then $|\overrightarrow{u} \times \overrightarrow{v}|$ is equal to