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 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

$A$ vector perpendicular to both of the vectors $i + j + k$ and $i + j$ is

Let $\overrightarrow{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{b}=\lambda \hat{j}+2 \hat{k}$, where $\lambda \in \mathbb{Z}$, be two vectors. Let $\overrightarrow{c}=\overrightarrow{a} \times \overrightarrow{b}$ and $\overrightarrow{d}$ be a vector of magnitude $2$ in the $yz$-plane. If $|\overrightarrow{c}|=\sqrt{53}$, then the maximum possible value of $(\overrightarrow{c} \cdot \overrightarrow{d})^2$ is equal to:

Let $m$ be a vector of magnitude $\sqrt{3}$ and perpendicular to the vectors $\hat{i}+\hat{j}$ and $\hat{j}-\hat{k}$. Let $n$ be another vector of magnitude $2\sqrt{6}$ and perpendicular to the vectors $2\hat{i}-\hat{j}$ and $\hat{j}+2\hat{k}$. The area (in sq. units) of the triangle formed with $m$ and $n$ as sides is

Let $F=2 \hat{i}+2 \hat{j}+5 \hat{k}$,$A=(1,2,5)$,$B=(-1,-2,-3)$ and $BA \times F=4 \hat{i}+6 \hat{j}+2 \lambda \hat{k}$,then $\lambda=$