If $\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}, \vec{b}=-3 \hat{i}+5 \hat{j}-4 \hat{k}$ and $\vec{c}=6 \hat{i}-4 \hat{j}+5 \hat{k}$,then $(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})=$

$\vec{a}=\hat{i}+\hat{j}-2 \hat{k}$, $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}+\hat{j}-\hat{k}$ are three vectors. If $\vec{d}$ is a normal to the plane of $\vec{a}$ and $\vec{b}$ and $\vec{d} \cdot \vec{c}=2$, then $|\vec{d}|=$

Let $\vec{a}=-\hat{i}+2\hat{j}+2\hat{k}$, $\vec{b}=8\hat{i}+7\hat{j}-3\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a}\times\vec{c}=\vec{b}$. If $\vec{c}\cdot(\hat{i}+\hat{j}+\hat{k})=4$, then $|\vec{a}+\vec{c}|^{2}$ is equal to:

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 $a \times (b \times c) = 0,$ then

If the magnitude of the vector product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of the vectors $2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\lambda \hat{i}+2 \hat{j}+3 \hat{k}$ is equal to $\sqrt{2}$,then the value of ' $\lambda$ ' is