The volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is $3 \text{ cubic units}$. The volume (in cubic units) of a tetrahedron with coterminous edges $(\vec{a} \times \vec{b}), (\vec{a} \times 2\vec{c}), (\vec{b} \times 2\vec{c})$ is:

$\vec{a}$ is a vector perpendicular to the plane containing non-zero vectors $\vec{b}$ and $\vec{c}$. If $\vec{a}, \vec{b}, \vec{c}$ are such that $|\vec{a}+\vec{b}+\vec{c}|=\sqrt{|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2}$, then $|(\vec{a} \times \vec{b}) \cdot \vec{c}|+|(\vec{a} \times \vec{b}) \times \vec{c}|=$

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 $3 p+q=$

Statement $1$: The vectors $\vec{a}, \vec{b}$ and $\vec{c}$ lie in the same plane if and only if $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$.
Statement $2$: The vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if $\vec{u} \cdot \vec{v} = 0$,where $\vec{u} \times \vec{v}$ is a vector perpendicular to the plane of $\vec{u}$ and $\vec{v}$.

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector, let us assume the vectors are $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$, $\vec{b} = \lambda \hat{i}+3 \hat{j}$, and we consider the standard basis vectors or a third vector to define coplanarity. However, if the question implies these two vectors are coplanar with the origin or a specific plane, we evaluate the scalar triple product. Given the standard interpretation of such problems, if $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ are coplanar with $\vec{c} = \hat{j}$, then the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 0$. Solving for $\lambda$ where $\vec{a} = (1, -3, 1)$, $\vec{b} = (\lambda, 3, 0)$, and $\vec{c} = (0, 1, 0)$:

If $a, b, c$ are any three coplanar unit vectors,then