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.

If $u = 2i + 2j - k$ and $v = 6i - 3j + 2k,$ then a unit vector perpendicular to both $u$ and $v$ is

Find a unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b},$ where $\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}.$

If non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular to each other,then the solution of the equation $\vec{r} \times \vec{a} = \vec{b}$ is:

Let $\vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}$,$\vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}$,$\vec{a} \cdot \vec{c}=7$,$2 \vec{b} \cdot \vec{c}+43=0$,and $\vec{a} \times \vec{c}=\vec{b} \times \vec{c}$. Then $|\vec{a} \cdot \vec{b}|$ is equal to

Two adjacent sides of a parallelogram are given by vectors $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$. Find the area of the parallelogram in square units.