If $\vec{a}=\alpha \hat{i}+\beta \hat{j}+3 \hat{k}$,$\vec{b}=\hat{j}+2 \hat{k}$,and $\vec{c}=3 \hat{i}+2 \hat{j}+\hat{k}$ are linearly dependent vectors and the magnitude of $\vec{a}$ is $\sqrt{14}$. If $\alpha$ and $\beta$ are integers,then $\alpha+\beta=$

If $a, b, c$ are vectors such that $[a, b, c] = 4$,then $[a \times b, b \times c, c \times a] = $

Let $\vec{c}$ be a vector coplanar with the unit vectors $\vec{a}$ and $\vec{b}$, and let $\vec{d}$ be the unit vector perpendicular to $\vec{a}$, $\vec{b}$, and $\vec{c}$. If $[\vec{a} \vec{b} \vec{d}] \vec{c} - [\vec{a} \vec{b} \vec{c}] \vec{d} = \hat{i} + 2\hat{j} + 2\hat{k}$ and the angle between $\vec{a}$ and $\vec{b}$ is $30^{\circ}$, then $|\vec{c}| =$

If $a = -3i + 7j + 5k$,$b = -3i + 7j - 3k$,and $c = 7i - 5j - 3k$ are the three coterminous edges of a parallelepiped,then its volume is

Let $a = i - k$, $b = xi + j + (1 - x)k$, and $c = yi + xj + (1 + x - y)k$. Then $[a\,b\,c]$ depends on

If the volume of the tetrahedron whose coterminous edges are given by the vectors $\vec{a} = -2\hat{i} + 3\hat{j} - 3\hat{k}$, $\vec{b} = 4\hat{i} + 5\hat{j} + (\lambda - 10)\hat{k}$, and $\vec{c} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ is $11$ cubic units, then the sum of the possible values of $\lambda$ is