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 $\lambda=$

The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8\hat{i} - 6\hat{j}$ and $3\hat{i} + 4\hat{j} - 12\hat{k}$ is:

Let $\bar{a}=4 \bar{i}+5 \bar{j}-\bar{k}$,$\bar{b}=\bar{i}-4 \bar{j}+5 \bar{k}$,$\bar{c}=3 \bar{i}+\bar{j}-\bar{k}$ and let $\bar{\alpha}$ be a vector perpendicular to both $\bar{a}$ and $\bar{b}$ such that $\bar{\alpha} \cdot \bar{c}=63$. Then $\bar{\alpha}=$

Let $\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$ be two vectors. If a vector perpendicular to both the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ has the magnitude $12$,then one such vector is

$A$ unit vector which is coplanar to vectors $i + j + 2k$ and $i + 2j + k$ and perpendicular to $i + j + k$ is

Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}$ be a vector such that $|\vec{c}-\vec{a}|=3$. If $\vec{p}=\vec{a} \times \vec{b}$,then the angle between $\vec{p}$ and $\vec{c}$ is $\frac{\pi}{6}$ and $|\vec{p} \times \vec{c}|=3$. Thus,$\vec{a} \cdot \vec{c}$ is equal to: