If $i, j, k$ are the unit vectors and mutually perpendicular,then $[i, k, j]$ is equal to

For non-zero vectors $\vec{a}, \vec{b}, \vec{c}$,the condition $|(\vec{a} \times \vec{b}) \cdot \vec{c}| = |\vec{a}||\vec{b}||\vec{c}|$ holds if and only if:

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 $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|^2 = \dots$

$a \cdot (b \times c)$ is equal to

The value of $a$ such that the volume of the parallelepiped formed by the vectors $i + aj + k$,$j + ak$,and $ai + k$ is minimum is:

The parallelepiped is determined by vectors $\vec{a} = -2\hat{i} + 5\hat{j} + 3\hat{k}$, $\vec{b} = \hat{i} + 3\hat{j} - 2\hat{k}$, and $\vec{c} = -3\hat{i} + \hat{j} + 4\hat{k}$. The altitude of the parallelepiped on the parallelogram base determined by vectors $\vec{b}$ and $\vec{c}$ is