If the vertices of $\Delta ABC$ are $A=(2,3,5), B=(-1,3,2), C=(3,5,-2)$,then the area of the $\Delta ABC$ (in sq. units) is

If $(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + \lambda \hat{j} + \mu \hat{k}) = \vec{0}$,then $\lambda$ and $\mu$ are respectively:

Let $|\vec{a}|=2, |\vec{b}|=3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{4}$. Then $|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2$ is equal to

Show that $(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})$.

If $a=2 \hat{i}+\hat{j}-3 \hat{k}$, $b=\hat{i}-2 \hat{j}+3 \hat{k}$, $c=-\hat{i}+\hat{j}-4 \hat{k}$ and $d=\hat{i}+\hat{j}+2 \hat{k}$, then $(a \times b) \times(c \times d)=$

If $\bar{a}=\hat{i}+\hat{j}+\hat{k}$ and $\bar{b}=\hat{j}-\hat{k}$,find a vector $\bar{c}$ such that $\bar{a} \times \bar{c}=\bar{b}$ and $\bar{a} \cdot \bar{c}=3$.