If the four points $2\vec{a} + 3\vec{b} - \vec{c}$,$\vec{a} - 2\vec{b} + 3\vec{c}$,$3\vec{a} + 4\vec{b} - 2\vec{c}$,and $\vec{a} - \lambda\vec{b} - 6\vec{c}$ are coplanar,find the value of $\lambda$.

$a$ is perpendicular to both $b$ and $c$. The angle between $b$ and $c$ is $\frac{2 \pi}{3}$. If $|a|=2$, $|b|=3$, and $|c|=4$, then $c \cdot (a \times b)$ is equal to (in $\sqrt{3}$)

If the points with position vectors $3i - 2j - k$,$2i + 3j - 4k$,$-i + j + 2k$,and $4i + 5j + \lambda k$ are coplanar,then $\lambda = \dots$

$[(\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c})] \cdot \vec{d} = \dots$

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 $\alpha + \beta + \gamma =$

Let the vectors $\vec{u} = (2+a+b) \hat{i}+(a+2 b+c) \hat{j}-(b+c) \hat{k}$,$\vec{v} = (1+b) \hat{i}+2 b \hat{j}-b \hat{k}$,and $\vec{w} = (2+b) \hat{i}+2 b \hat{j}+(1-b) \hat{k}$ where $a, b, c \in \mathbb{R}$ be co-planar. Then which of the following is true?