Let $\overrightarrow{a}=\hat{i}-2 \hat{j}$,$\overrightarrow{b}=2 \hat{j}+3 \hat{k}$,$\overrightarrow{c}=p\hat{i}+q \hat{j}$ and $\overrightarrow{d}=p \hat{j}-q \hat{k}$ be four vectors. If $(\vec{a} \times \vec{b}) \cdot \vec{c}=3=(\vec{a} \times \vec{b}) \cdot \vec{d}$,then $3 p+q=$

If $a, b$ and $c$ are three non-coplanar vectors and $p, q$ and $r$ are 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

If $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = r\hat{i} + \hat{j} + (2r - 1)\hat{k}$ are three vectors such that $\vec{c}$ is parallel to the plane of $\vec{a}$ and $\vec{b}$,then $r$ is equal to

The volume of the parallelepiped whose edges are represented by $-12i + \alpha k$,$3j - k$,and $2i + j - 15k$ is $546$. Then $\alpha = $

If $\alpha = 2i + 3j - k$,$\beta = -i + 2j - 4k$,and $\gamma = i + j + k$,then $(\alpha \times \beta) \cdot (\alpha \times \gamma)$ is equal to

If the volume of the tetrahedron whose coterminous edges are given by the vectors $\vec{a} = -2\hat{i} + 3\hat{j} - 3\hat{k}$, $\vec{b} = 4\hat{i} + 5\hat{j} + (\lambda - 10)\hat{k}$, and $\vec{c} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ is $11$ cubic units, then the sum of the possible values of $\lambda$ is