The area of the triangle with vertices $A(1, 1, 2)$, $B(2, 3, 5)$ and $C(1, 5, 5)$ is . . . . . . .

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})$. If $|\vec{a}| = 1, |\vec{b}| = 4, |\vec{c}| = 2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^{\circ}$, then $|\vec{a} \cdot \vec{c}|$ is:

If a non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{j}-\hat{k}$ and $3\hat{j}-2\hat{k}$ and the plane determined by the vectors $2\hat{i}+3\hat{j}$ and $\hat{i}-3\hat{j}$,then the angle between the vectors $\vec{a}$ and $\hat{i}+\hat{j}+\hat{k}$ is

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{j}-\hat{k}$ are given vectors, then a vector $\vec{b}$ satisfying the equations $\vec{a} \times \vec{b}=\vec{c}$ and $\vec{a} \cdot \vec{b}=3$ is

$\vec{a}=\hat{i}+\hat{j}-2 \hat{k}$, $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ 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 $\bar{a}$ and $\bar{b}$ be two vectors such that $|\bar{a}|=1$,$|\bar{b}|=4$,and $\bar{a} \cdot \bar{b}=2$. If $\bar{c}=(2 \bar{a} \times \bar{b})-3 \bar{b}$,then the angle between $\bar{b}$ and $\bar{c}$ is