The points $A(4, 5, 1)$,$B(0, -1, -1)$,$C(3, 9, 4)$,and $D(-4, 4, 4)$ are:

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors,then $\left| {\begin{array}{*{20}{c}} {\vec{a} \cdot \vec{a}} & {\vec{a} \cdot \vec{b}} & {\vec{a} \cdot \vec{c}} \\ {\vec{b} \cdot \vec{a}} & {\vec{b} \cdot \vec{b}} & {\vec{b} \cdot \vec{c}} \\ {\vec{c} \cdot \vec{a}} & {\vec{c} \cdot \vec{b}} & {\vec{c} \cdot \vec{c}} \end{array}} \right| = \dots$

Let $\overrightarrow{A} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{B} = \hat{i}$,and $\overrightarrow{C} = C_1\hat{i} + C_2\hat{j} + C_3\hat{k}$. If $C_2 = -1$ and $C_3 = 1$,then to make the three vectors coplanar:

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 $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$,$\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$. If $\vec{c} = x\vec{a} + y\vec{b} + z(\vec{a} \times \vec{b})$,then the value of $x + y + z$ is:

If $a=2u+3v+7w$, $b=u+v-2w$ and $c=-u-2v-3w$, then $\left|\frac{[u, v, w]}{[a, b, c]}\right|(a+b+c) = $