$[i, k, j] + [k, j, i] + [j, k, i]$

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-zero and non-coplanar vectors such that $(\vec{a} + \lambda \vec{b}) \cdot [(\vec{b} + 3\vec{c}) \times (\vec{c} - 4\vec{a})] = 0$,then $\lambda$ is equal to

Let $\alpha \in \mathbb{R}$ and the three vectors $\vec{a} = \alpha \hat{i} + \hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + \hat{j} - \alpha \hat{k}$,and $\vec{c} = \alpha \hat{i} - 2\hat{j} + 3\hat{k}$. Then the set $S = \{ \alpha : \vec{a}, \vec{b}, \text{ and } \vec{c} \text{ are coplanar} \}$

If $\overline{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $\overline{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$,then the value of $(2 \bar{a}-\bar{b}) \cdot [(\bar{a} \times \bar{b}) \times (\bar{a}+2 \bar{b})] = $

If $\vec{w} = \alpha (\vec{a} \times \vec{b}) + \beta (\vec{b} \times \vec{c}) + \gamma (\vec{c} \times \vec{a})$,$[\vec{a}, \vec{b}, \vec{c}] = 2$ and $\vec{w} \cdot (\vec{a} + \vec{b} + \vec{c}) = 8$,then $\alpha + \beta + \gamma =$