If $\overline{a}=\hat{i}-\hat{k}$,$\overline{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\overline{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$,then $\overline{a} \cdot(\overline{b} \times \overline{c})$ depends on

The position vectors of the points $A, B, C$ and $D$ are $3 \hat{i}-2 \hat{j}-\hat{k}, 2 \hat{i}-3 \hat{j}+2 \hat{k}, \hat{i}-\hat{j}+2 \hat{k}$ and $4 \hat{i}-\hat{j}-\lambda \hat{k}$ respectively. If the points $A, B, C$ and $D$ lie on a plane, the value of $\lambda$ is

Let the vectors $\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}$,$\vec{b}=(1-t) \hat{i}+(1+t) \hat{j}+2 \hat{k}$ and $\vec{c}=\hat{i}-t \hat{j}+\hat{k}$,$t \in R$ be such that for $\alpha, \beta, \gamma \in R$,$\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\vec{0} \Rightarrow \alpha=\beta=\gamma=0$. Then,the set of all values of $t$ is:

Let $a=\hat{i}+\hat{j}$, $b=\hat{j}+\hat{k}$ and $c=\hat{i}+\hat{k}$. If $d$ is a unit vector such that $a \cdot d=0$ and $b \cdot(c \times d)=0$, then $d=$

If four distinct points with position vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are coplanar,then $[\vec{a} \vec{b} \vec{c}]$ is equal to

Let $\vec{a} = \hat{i} - \hat{k}$,$\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}$,and $\vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$. Then the scalar triple product $[\vec{a} \, \vec{b} \, \vec{c}]$ depends on: