If $\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}+\hat{k}$,and $\vec{c}=3 \hat{i}-\hat{j}+2 \hat{k}$,then $\vec{a} \cdot(\vec{b} \times \vec{c})=$ . . . . . . .

If the vectors $2i - 3j + 4k$, $i + 2j - k$ and $xi - j + 2k$ are coplanar, then $x = $

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 $\vec{a}$ is a unit vector perpendicular to both $\vec{b}$ and $\vec{c}$. If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$, then find the value of $\left| \begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{matrix} \right|^2$.

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+2\hat{k}$,$\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$ and $\vec{c}$ is a linear combination of $\vec{a}$ and $\vec{b}$,then the value of $x$ is:

The vectors $i + 2j + 3k$,$\lambda i + 4j + 7k$,and $-3i - 2j - 5k$ are collinear if $\lambda$ equals:

If $\bar{u}, \bar{v},$ and $\bar{w}$ are three non-coplanar vectors,then $(\bar{u} + \bar{v} - \bar{w}) \cdot (\bar{u} - \bar{v}) \times (\bar{v} - \bar{w}) = \dots$