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:

Let $\vec{a} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = |\vec{c}|$, $|\vec{c} - \vec{a}| = 2\sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $60^\circ$, then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is equal to

If $\overline{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\overline{OB} = 4\hat{i} + \hat{k}$ are the position vectors of the points $A$ and $B$, then the position vector of a point on the line passing through $B$ and parallel to the vector $\overline{OA} \times \overline{OB}$ which is at a distance of $\sqrt{189}$ units from $B$ is

Let $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\overline{b}=\hat{i}+\hat{j}$. If $\overline{c}$ is a vector such that $\overline{a} \cdot \overline{c}=|\overline{c}|$,$|\overline{c}-\overline{a}|=2 \sqrt{2}$,and the angle between $(\overline{a} \times \overline{b})$ and $\overline{c}$ is $30^{\circ}$,then the value of $|(\overline{a} \times \overline{b}) \times \overline{c}|$ is equal to

If $\vec{a}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-3\hat{k}$,then the unit vector perpendicular to both $\vec{p}=\vec{a}-\vec{b}$ and $\vec{q}=\vec{a}+\vec{b}$ is . . . . . . .

If $\overrightarrow{a}=2 \hat{i}-5 \hat{j}+8 \hat{k}$ and $\overrightarrow{b}=7 \hat{i}-5 \hat{j}+3 \hat{k}$ are two vectors and $(2 \overrightarrow{a}-3 \overrightarrow{b}) \times(4 \overrightarrow{a}+\overrightarrow{b})=x \hat{i}+y \hat{j}+z \hat{k}$,then $x+y+z=$