If the volume of a tetrahedron,whose vertices are $A(1, 2, 3)$,$B(-3, -1, 1)$,$C(2, 1, 3)$,and $D(-1, 2, x)$ is $\frac{11}{6}$ cubic units,then the value of $x$ is:

If $|\vec{a}| = 4, |\vec{b}| = 3$ and $\vec{a} \cdot \vec{b} = 8$, then the scalar triple product $[\vec{a} \quad \vec{a} + \vec{b} \quad \vec{a} \times \vec{b}]$ is equal to:

If $a = 3i - 2j + 2k$,$b = 6i + 4j - 2k$ and $c = 3i - 2j - 4k$,then $a \cdot (b \times c)$ is

Let $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$,$\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$. If $\vec{c} = x\vec{a} + y\vec{b} + z(\vec{a} \times \vec{b})$,then the value of $x + y + z$ is:

$(\hat{i} \times \hat{j}) \cdot [(\hat{j} \times \hat{k}) \times (\hat{k} \times \hat{i})]$

$\vec{a}$ is a vector perpendicular to the plane containing non-zero vectors $\vec{b}$ and $\vec{c}$. If $\vec{a}, \vec{b}, \vec{c}$ are such that $|\vec{a}+\vec{b}+\vec{c}|=\sqrt{|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2}$, then $|(\vec{a} \times \vec{b}) \cdot \vec{c}|+|(\vec{a} \times \vec{b}) \times \vec{c}|=$