The points $A(5, -1, 1)$,$B(7, -4, 7)$,$C(1, -6, 10)$,and $D(-1, -3, 4)$ are vertices of a:

If $\vec{a}=\hat{i}+2\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+4\hat{k}$,and $\vec{c}=\hat{i}+\hat{j}+\hat{k}$ are such that $\vec{a}+\lambda\vec{b}$ is perpendicular to $\vec{c}$,then the value of $\lambda$ is

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

$a$ and $b$ are non-collinear vectors,$|a|=2 \sqrt{2}$,$|b|=3$ and the angle between $a$ and $b$ is $45^{\circ}$. Then,the lengths of the diagonals of the parallelogram whose adjacent sides are represented by the vectors $5a+2b$ and $a-3b$ are

If $\bar{a}$ and $\bar{b}$ are unit vectors such that $|\bar{a}+\bar{b}|=\sqrt{3}$,then the angle between $\bar{a}$ and $\bar{b}$ is

If $|\vec{a}| = \sqrt{27}$,$|\vec{b}| = 7$ and $|\vec{a} \times \vec{b}| = 35$,then $\vec{a} \cdot \vec{b}$ is equal to