If the position vectors of the vertices of a $\triangle ABC$ are $\vec{OA} = 3\hat{i} + \hat{j} + 2\hat{k}$,$\vec{OB} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{OC} = 2\hat{i} + 3\hat{j} + \hat{k}$,then the length of the altitude of $\triangle ABC$ drawn from $A$ is

If the vertices of a $\triangle ABC$ are $A=(2,3,5)$,$B=(-1,3,2)$,and $C=(3,5,-2)$,then the area of the $\triangle ABC$ (in sq. units) 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 $\frac{2 \pi}{3}$,then $|(\overline{a} \times \overline{b}) \times \overline{c}|=$

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})$. If $|\vec{a}| = 1, |\vec{b}| = 4, |\vec{c}| = 2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^{\circ}$, then $|\vec{a} \cdot \vec{c}|$ is:

If $a=\hat{i}+2 \hat{j}+3 \hat{k}$,$b=-\hat{i}+2 \hat{j}+\hat{k}$,$c=\hat{i}+2 \hat{j}-2 \hat{k}$,$n$ is perpendicular to both $a$ and $b$,and $\theta$ is the angle between $c$ and $n$,then $\sin \theta=$

If $1, 2, 3$ and $-1, 0, 1$ are the direction ratios of the rays $OA$ and $OB$ respectively, then the direction cosines of a normal to the plane $AOB$ are