The area of the parallelogram,whose diagonals are $\vec{d}_1 = 2 \hat{i} - \hat{j} + \hat{k}$ and $\vec{d}_2 = \hat{i} + 3 \hat{j} - \hat{k}$,is equal to

If $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$,$\vec{b}=3(\hat{i}-\hat{j}+\hat{k})$ and $\vec{c}$ is a vector such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3$,then $\vec{a} \cdot(\vec{c} \times \vec{b}-\vec{b}-\vec{c})=$

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:

If the position vectors of the vertices of $\triangle ABC$ are $3\hat{i}+4\hat{j}-\hat{k}$,$\hat{i}+3\hat{j}+\hat{k}$,and $5(\hat{i}+\hat{j}+\hat{k})$ respectively,then the magnitude of the altitude from $A$ onto the side $BC$ is

If $\bar{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$,$\bar{b}=\hat{i}-2 \hat{j}-2 \hat{k}$,$\bar{c}=-\hat{i}+4 \hat{j}+3 \hat{k}$ and if $\bar{d}$ is a vector perpendicular to both $\bar{b}$ and $\bar{c}$,and $\bar{a} \cdot \bar{d}=18$,then $|\bar{a} \times \bar{d}|^2=$

If $(\bar{i}+\bar{j}+\bar{k})$,$(\bar{i}+2\bar{j}+3\bar{k})$ and $(2\bar{i}-\bar{j}+\bar{k})$ are the position vectors of the vertices $A$,$B$ and $C$ of $\triangle ABC$ respectively,then the vector equation of the altitude through $A$ is