$(a+b) \cdot(b+c) \times(a+b+c)$ is equal to

The volume of the tetrahedron formed by the vectors $\vec{a}, \vec{b}, \vec{c}$ is $3$. Then the volume of the parallelepiped formed by the coterminous edges $\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}$ is:

$a \cdot (b \times c)$ is equal to

If the scalar triple product of the vectors $-3 \hat{i}+7 \hat{j}-3 \hat{k}$,$3 \hat{i}-7 \hat{j}+\lambda \hat{k}$ and $7 \hat{i}-5 \hat{j}-3 \hat{k}$ is $272$,then $\lambda = \ldots$

Consider the four points $A(1, -2, -1)$, $B(4, 0, -3)$, $C(1, 2, -1)$, and $D(2, -4, -5)$ in space. If $\vec{b} = \vec{AB}$, $\vec{c} = \vec{AC}$, and $\vec{d} = \vec{AD}$, then find the value of $\frac{[\vec{b} \times \vec{c}, \vec{c} \times \vec{d}, \vec{d} \times \vec{b}]}{[\vec{b}+\vec{c}, \vec{c}+\vec{d}, \vec{d}+\vec{b}]}$.

If $\bar{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \bar{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$,and $\bar{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$,and $[3 \bar{a}+\bar{b} \quad 3 \bar{b}+\bar{c} \quad 3 \bar{c}+\bar{a}] = \lambda \begin{vmatrix} \bar{a} \cdot \hat{i} & \bar{a} \cdot \hat{j} & \bar{a} \cdot \hat{k} \\ \bar{b} \cdot \hat{i} & \bar{b} \cdot \hat{j} & \bar{b} \cdot \hat{k} \\ \bar{c} \cdot \hat{i} & \bar{c} \cdot \hat{j} & \bar{c} \cdot \hat{k} \end{vmatrix}$,then the value of $\lambda$ is: