Find $|\vec{a} \times \vec{b}|,$ if $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k}$.

Vectors $\vec{p}=a \hat{i}+b \hat{j}+c \hat{k}$, $\vec{q}=d \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}$ form a triangle $ABC$ such that $\vec{p}=\vec{q}+\vec{r}$. If the area of $\triangle ABC$ is $5 \sqrt{6}$ sq. units, then the sum of the absolute values of $a, b, c$ is

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

Let $\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$,for some real $x$. Then $|\vec{a} \times \vec{b}| = r$ is possible if

The diagonals of a parallelogram are $\vec{d_1} = \hat{j} + \hat{k}$ and $\vec{d_2} = \hat{i} + \hat{j}$. The area of the parallelogram is . . . . . . sq. units.

Let $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$, $\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}$ and a vector $\vec{c}$ be such that $2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0}$. If $\vec{a} \cdot \vec{c} = 15$, then $\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k})$ is equal to: