The value of $a$,so that the volume of the parallelepiped formed by $\hat{i} + a \hat{j} + \hat{k}$,$\hat{j} + a \hat{k}$,and $a \hat{i} + \hat{k}$ becomes minimum is

If the points with position vectors $\hat{i}-2 \hat{j}+3 \hat{k}$, $2 \hat{i}+3 \hat{j}-4 \hat{k}$, $-3 \hat{i}+\hat{j}-5 \hat{k}$, and $a \hat{i}-2 \hat{j}+4 \hat{k}$ are coplanar, then $a=$

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 $\vec a = 3\vec j + 4\vec k$,$\vec b = 2\vec i + \vec k$ and $\vec c$,$\vec d$ are respectively the components of $\vec a$ parallel and perpendicular to $\vec b$,then the value of the scalar triple product $\left[ {(\vec a \times \vec c) \times (\vec c \times \vec d), (\vec c \times \vec d) \times (\vec d \times \vec a), (\vec d \times \vec a) \times (\vec a \times \vec c)} \right]$ is equal to:

Let $\vec{a}=2 \hat{i}+7 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}+5 \hat{k}$ and $\vec{c}=\hat{i}-\hat{j}+2 \hat{k}$. Let $\vec{d}$ be a vector which is perpendicular to both $\vec{a}$ and $\vec{b}$,and $\vec{c} \cdot \vec{d}=12$. Then $(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d})$ is equal to $........$.

Unit vectors $a, b$ and $c$ are coplanar. $A$ unit vector $d$ is perpendicular to them. If $(a \times b) \times (c \times d) = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$ and the angle between $a$ and $b$ is $30^\circ$,then $c$ is: