$\vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=\hat{i}+2 \hat{j}-\hat{k}$ are two vectors and $\vec{a}$ is a unit vector such that $\cos (\vec{a}, \vec{b} \times \vec{c})=\sqrt{\frac{2}{3}}$. Then $|\vec{a} \times(\vec{b} \times \vec{c})|=$

Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = 3\hat{i} - \hat{j} + 5\hat{k}$, and $\vec{c} = \hat{i} - 4\hat{j} - 2\hat{k}$ be three vectors. Let $\vec{r}$ be a vector perpendicular to both $\vec{b}$ and $\vec{c}$, and $\vec{r} \cdot \vec{a} = 11$. Then the vector among the following that is perpendicular to $\vec{r}$ is:

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

Let $L_1: \frac{x+1}{3}=\frac{y+2}{2}=\frac{z+1}{1}$ and $L_2: \frac{x-2}{2}=\frac{y+2}{1}=\frac{z-3}{3}$ be the given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is

The area of a triangle whose vertices are $A(1, -1, 2)$,$B(2, 1, -1)$ and $C(3, -1, 2)$ is

The area of a parallelogram whose adjacent sides are $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = -\hat{j} - 2\hat{k}$ is . . . . . . sq. units.