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 parallelopiped whose coterminous edges are $\hat{j}+\hat{k}$, $\hat{i}+\hat{k}$, and $\hat{i}+\hat{j}$ is

Let $\vec{u} = a\hat{i} + b\hat{j} + c\hat{k}$,$\vec{v} = b\hat{i} + c\hat{j} + a\hat{k}$,and $\vec{w} = c\hat{i} + a\hat{j} + b\hat{k}$. If $[\vec{u} \, \vec{v} \, \vec{w}] = 0$ and $\vec{w} = \lambda \vec{x} + \mu \vec{y}$ where $(a + b + c) \neq 0$ and $\lambda, \mu \neq 0$,then the vectors $\vec{x}, \vec{y}, \vec{u}, \vec{v}, \vec{w}$ are:

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero non-coplanar vectors. Let the position vectors of four points $A, B, C$ and $D$ be $\vec{a}-\vec{b}+\vec{c}$,$\lambda \vec{a}-3 \vec{b}+4 \vec{c}$,$-\vec{a}+2 \vec{b}-3 \vec{c}$ and $2 \vec{a}-4 \vec{b}+6 \vec{c}$ respectively. If $\overrightarrow{AB}$,$\overrightarrow{AC}$ and $\overrightarrow{AD}$ are coplanar,then $\lambda$ is :

If $[\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}] = \lambda [\vec{a}, \vec{b}, \vec{c}]^2$,then $\lambda$ is equal to:

The vectors $2 \hat{i}-3 \hat{j}+\hat{k}, \hat{i}-2 \hat{j}+3 \hat{k}$ and $3 \hat{i}+\hat{j}-2 \hat{k}$