Let $A=(\alpha, 1, 2\alpha)$,$B=(3, 1, 2)$ and $C=4\hat{i}-\hat{j}+3\hat{k}$. If $AB \times C = 6\hat{i}+9\hat{j}-5\hat{k}$,then $\alpha^2+\alpha+5=$

$A$ unit vector perpendicular to the plane determined by the points $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$ is:

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

If $\overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b}$,$\overrightarrow{v}=\overrightarrow{a}+\overrightarrow{b}$,$|\overrightarrow{a}|=|\overrightarrow{b}|=2$,then $|\overrightarrow{u} \times \overrightarrow{v}|$ is equal to

Let $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+2 \hat{k}$. Then the number of vectors $\vec{b}$ such that $\vec{b} \times \vec{c}=\vec{a}$ and $|\vec{b}| \in\{1, 2, \ldots, 10\}$ is

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