If the vectors $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$,$\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}$,and $\vec{c}=3 \hat{i}+p \hat{j}+5 \hat{k}$ are coplanar,then $p=$

The volume of the tetrahedron having the edges $\hat{i}+2\hat{j}-\hat{k}$, $\hat{i}+\hat{j}+\hat{k}$, and $\hat{i}-\hat{j}+\lambda\hat{k}$ as coterminous edges is $\frac{2}{3}$ cubic units. Then $\lambda$ equals:

The volume of the tetrahedron with $\hat{i}-\lambda \hat{j}+\hat{k}$, $\lambda \hat{i}-\hat{j}-\hat{k}$ and $\hat{i}+\hat{j}+\lambda \hat{k}$ as coterminous edges is $2$. If $\lambda$ is an integer, then $|\lambda \hat{i}-3 \lambda \hat{j}+3 \hat{k}|=$

If $a, b, c, d$ are $4$ vectors such that $a \cdot b = 0$, $|a \times c| = |a||c|$, and $|a \times d| = |a||d|$, then $[b c d] = $

If $a=2u+3v+7w$, $b=u+v-2w$ and $c=-u-2v-3w$, then $\left|\frac{[u, v, w]}{[a, b, c]}\right|(a+b+c) = $

If $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar vectors and $p, q$ are real numbers,then the equality $[3\vec{u}, p\vec{v}, p\vec{w}] - [p\vec{v}, \vec{w}, q\vec{u}] - [2\vec{w}, q\vec{v}, q\vec{u}] = 0$ holds for: