Find the unit vector perpendicular to the plane containing the points $(1, -1, 2), (2, 0, -1),$ and $(0, 2, 1)$.

$A$ unit vector perpendicular to the plane containing the vectors $\hat{i}+2\hat{j}+\hat{k}$ and $-2\hat{i}+\hat{j}+3\hat{k}$ is

If $\vec{a}$ is a vector such that $\vec{a} \times \hat{i}=\hat{j}+\hat{k}$ and $\vec{a} \cdot \hat{i}=1$, then the equation of the line passing through the point $\hat{i}+\hat{j}+\hat{k}$ and parallel to $\vec{a}$ is

$A$ vector with magnitude of $3$ units,which is perpendicular to each of the vectors $\vec{a}=3 \hat{i}+\hat{j}-4 \hat{k}$ and $\vec{b}=6 \hat{i}+5 \hat{j}-2 \hat{k}$,is given by

Let $\hat{\alpha}, \hat{\beta}, \hat{\gamma}$ be three unit vectors such that $\hat{\alpha} \times (\hat{\beta} \times \hat{\gamma}) = \frac{1}{2}(\hat{\beta} + \hat{\gamma})$. If $\hat{\beta}$ is not parallel to $\hat{\gamma}$, then the angle between $\hat{\alpha}$ and $\hat{\beta}$ is

Find the area of a parallelogram whose adjacent sides are given by the vectors $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.