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 $\lambda = \ldots$

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

If the origin $O(0,0,0)$ and the points $P(2,3,4)$,$Q(1,2,3)$,and $R(x, y, z)$ are co-planar,then:

If $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$ are coterminous edges of a parallelepiped,then its volume is

Let the position vectors of the points $A, B, C$ and $D$ be $5\hat{i}+5\hat{j}+2\lambda\hat{k}$,$\hat{i}+2\hat{j}+3\hat{k}$,$-2\hat{i}+\lambda\hat{j}+4\hat{k}$ and $-\hat{i}+5\hat{j}+6\hat{k}$. Let the set $S = \{\lambda \in \mathbb{R} : \text{The points } A, B, C \text{ and } D \text{ are coplanar}\}$. Then $\sum_{\lambda \in S}(\lambda+2)^2$ is equal to

Three vectors $\hat{i}-\hat{k}$,$\lambda \hat{i}+\hat{j}+(1-\lambda) \hat{k}$,and $\mu \hat{i}+\lambda \hat{j}+(1+\lambda-\mu) \hat{k}$ represent the coterminous edges of a parallelepiped. The volume of the parallelepiped depends on: