The number of integral values of $p$ for which the vectors $(p+1) \hat{i} - 3 \hat{j} + p \hat{k}$,$p \hat{i} + (p+1) \hat{j} - 3 \hat{k}$,and $-3 \hat{i} + p \hat{j} + (p+1) \hat{k}$ are linearly dependent 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

The volume of the parallelepiped having vertices at $O \equiv (0,0,0)$, $A \equiv (2,-2,1)$, $B \equiv (5,-4,4)$, and $C \equiv (1,-2,4)$ is:

If the volume of a parallelepiped,whose coterminous edges are given by the vectors $\overrightarrow{a} = \hat{i} + \hat{j} + n\hat{k}$,$\overrightarrow{b} = 2\hat{i} + 4\hat{j} - n\hat{k}$,and $\overrightarrow{c} = \hat{i} + n\hat{j} + 3\hat{k}$ $(n \geq 0)$,is $158$ cubic units,then which of the following is true?

If $A(-1, 2, 3)$,$B(3, -2, 1)$,$C(2, 1, 3)$ and $D(-1, -2, 4)$ are the vertices of a tetrahedron,then its volume is

The volume of the tetrahedron having vertices $(1, -6, 10)$,$(-1, -3, 7)$,$(5, -1, \lambda)$ and $(7, -4, 7)$ is $11 \text{ cubic units}$. Then $\lambda = $