The cosine of the angle $A$ of the triangle with vertices $A(1, -1, 2)$,$B(6, 11, 2)$,and $C(1, 2, 6)$ is:

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be vectors of equal magnitude such that the angle between $\vec{a}$ and $\vec{b}$ is $\alpha$, $\vec{b}$ and $\vec{c}$ is $\beta$, and $\vec{c}$ and $\vec{a}$ is $\gamma$. Then the minimum value of $\cos \alpha + \cos \beta + \cos \gamma$ is

If $a \cdot b = a \cdot c$,$a \times b = a \times c$ and $a \neq 0$,then

Let $\vec{c}$ be the projection vector of $\vec{b}=\lambda \hat{i}+4 \hat{k}, \lambda>0$,on the vector $\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k}$. If $|\vec{a}+\vec{c}|=7$,then the area of the parallelogram formed by the vectors $\vec{b}$ and $\vec{c}$ is . . . . . . .

If $P=(0,1,2)$, $Q=(4,-2,1)$, and $O=(0,0,0)$, then $\angle POQ$ is equal to

Consider two vectors $\overrightarrow{u} = 3\hat{i} - \hat{j}$ and $\overrightarrow{v} = 2\hat{i} + \hat{j} - \lambda\hat{k}$,where $\lambda > 0$. The angle between them is given by $\cos^{-1}\left(\frac{\sqrt{5}}{2\sqrt{7}}\right)$. Let $\vec{v} = \vec{v}_1 + \vec{v}_2$,where $\vec{v}_1$ is parallel to $\overrightarrow{u}$ and $\vec{v}_2$ is perpendicular to $\overrightarrow{u}$. Then the value $|\vec{v}_1|^2 + |\vec{v}_2|^2$ is equal to