The volume of the tetrahedron whose coterminus edges are represented by $\bar{a}=-12 \hat{i}+p \hat{k}$,$\bar{b}=3 \hat{j}-\hat{k}$,and $\bar{c}=2 \hat{i}+\hat{j}-15 \hat{k}$ is $570$ cubic units. Then,$p=$

Let $\vec{OD} = \hat{i} + 2\hat{j} + 6\hat{k}$ and $\vec{CB} = -3\hat{i} - 2\hat{k}$ be the diagonals of the parallelogram $OBDC$. If $\vec{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$, then the volume of the parallelepiped determined by vectors $\vec{OA}, \vec{OB}$, and $\vec{OC}$ (in cubic units) is:

If $\vec{r}$ is a vector perpendicular to both the vectors $2 \hat{i}+3 \hat{j}-4 \hat{k}$ and $3 \hat{i}-\hat{j}+\hat{k}$ and satisfies $\vec{r} \cdot(3 \hat{i}-3 \hat{j}+4 \hat{k})=5$, then $|\vec{r}|=$

If the points having the position vectors $-\hat{i}+4 \hat{j}-4 \hat{k}$,$3 \hat{i}+2 \hat{j}-5 \hat{k}$,$-3 \hat{i}+8 \hat{j}-5 \hat{k}$ and $-3 \hat{i}+2 \hat{j}+\lambda \hat{k}$ are coplanar,then $\lambda=$

Let $a, b$ and $c$ be three non-coplanar vectors and let $p, q$ and $r$ be the vectors defined by $p=\frac{b \times c}{[a b c]}, q=\frac{c \times a}{[a b c]}, r=\frac{a \times b}{[a b c]}$. Then, $(a+b) \cdot p+(b+c) \cdot q+(c+a) \cdot r$ is equal to

For three vectors $a, b, c$,the value of $[a \times b, b \times c, c \times a]$ is equal to: