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

Let a vector $\vec{a}$ be coplanar with vectors $\vec{b}=2 \hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}+\hat{k}$. If $\vec{a}$ is perpendicular to $\vec{d}=3 \hat{i}+2 \hat{j}+6 \hat{k}$,and $|\vec{a}|=\sqrt{10}$. Then a possible value of $[\vec{a} \vec{b} \vec{c}]+[\vec{a} \vec{b} \vec{d}]+[\vec{a} \vec{c} \vec{d}]$ is equal to:

If the volume of a tetrahedron whose conterminous edges are $\overline{a}+\overline{b}, \overline{b}+\overline{c}, \overline{c}+\overline{a}$ is $24$ cubic units,then the volume of the parallelepiped whose coterminous edges are $\overline{a}, \overline{b}, \overline{c}$ is

If $a = i + j + k$,$b = 4i + 3j + 4k$ and $c = i + \alpha j + \beta k$ are linearly dependent vectors and $|c| = \sqrt{3}$,then

If the vectors $ai + j + k$,$i + bj + k$,and $i + j + ck$ $(a \ne 1, b \ne 1, c \ne 1)$ are coplanar,then the value of $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = $

If $a = 2i + j - k$,$b = i + 2j + k$,and $c = i - j + 2k$,then $a \cdot (b \times c) = \dots$