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:

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$,$\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$,and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$ be three non-zero vectors such that $\vec{c}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$. If the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,then $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|^2 = \dots$

Let $OA, OB, OC$ be the co-terminal edges of a rectangular parallelopiped of volume $V$ and let $P$ be the vertex opposite to $O$. Then, $[\overrightarrow{AP} \overrightarrow{BP} \overrightarrow{CP}]$ is equal to

If the vectors $a$,$b$,and $c$ are coplanar,then $\left|\begin{array}{ccc}a & b & c \\ a \cdot a & a \cdot b & a \cdot c \\ b \cdot a & b \cdot b & b \cdot c\end{array}\right|$ is equal to

Let $\vec{a} = \hat{i} - \hat{k}$,$\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}$,and $\vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$. Then the scalar triple product $[\vec{a} \, \vec{b} \, \vec{c}]$ depends on:

Let $a, b$ and $c$ be distinct non-negative numbers. If the vectors $a\hat{i} + a\hat{j} + c\hat{k}$,$\hat{i} + \hat{k}$ and $c\hat{i} + c\hat{j} + b\hat{k}$ are coplanar,then $c = \dots$