Find the area of a triangle having the points $A(1, 1, 1)$,$B(1, 2, 3)$,and $C(2, 3, 1)$ as its vertices.

Let $\vec{u} = \hat{i} + \hat{j}$,$\vec{v} = \hat{i} - \hat{j}$,and $\vec{w} = \hat{i} + 2\hat{j} + 3\hat{k}$. If $\hat{n}$ is a unit vector such that $\vec{u} \cdot \hat{n} = 0$ and $\vec{v} \cdot \hat{n} = 0$,then $|\vec{w} \cdot \hat{n}| = ....$

$(a - b) \times (a + b) = $

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{c}=\hat{j}-\hat{k}$,$\vec{a} \times \vec{b}=\vec{c}$ and $\vec{a} \cdot \vec{b}=1$,then $\vec{b}$ is equal to:

$\vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ 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 $|\vec{a} \times(\vec{b} \times \vec{c})|=$

The area of a parallelogram whose adjacent sides are given by the vectors $i + 2j + 3k$ and $-3i - 2j + k$ (in square units) is