Let $a = i - k$, $b = xi + j + (1 - x)k$, and $c = yi + xj + (1 + x - y)k$. Then $[a\,b\,c]$ depends on

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:

If $a, b, c$ are any three vectors and their reciprocal vectors are $a^{-1}, b^{-1}, c^{-1}$ such that $[a, b, c] \neq 0$,then $[a^{-1}, b^{-1}, c^{-1}]$ is equal to:

If $\vec{OA}=6 \hat{i}+3 \hat{j}-4 \hat{k}$, $\vec{OB}=2 \hat{j}+\hat{k}$, and $\vec{OC}=5 \hat{i}-\hat{j}+2 \hat{k}$ are the coterminous edges of a parallelepiped, then the height of the parallelepiped drawn from the vertex $A$ is

Let $\overrightarrow{u}, \overrightarrow{v}$ and $\overrightarrow{w}$ be vectors in three-dimensional space,where $\overrightarrow{u}$ and $\overrightarrow{v}$ are unit vectors which are not perpendicular to each other and $\overrightarrow{u} \cdot \overrightarrow{w}=1, \overrightarrow{v} \cdot \overrightarrow{w}=1, \overrightarrow{w} \cdot \overrightarrow{w}=4$. If the volume of the parallelepiped,whose adjacent sides are represented by the vectors $\overrightarrow{u}, \overrightarrow{v}$ and $\overrightarrow{w}$,is $\sqrt{2}$,then the value of $|3\vec{u}+5\vec{v}|$ is.

Let $\vec{a}=2 \hat{i}+7 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}+5 \hat{k}$ and $\vec{c}=\hat{i}-\hat{j}+2 \hat{k}$. Let $\vec{d}$ be a vector which is perpendicular to both $\vec{a}$ and $\vec{b}$,and $\vec{c} \cdot \vec{d}=12$. Then $(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d})$ is equal to $........$.