Let $\mathbf{a}$ and $\mathbf{b}$ be non-collinear vectors. If the vectors $(\lambda-1) \mathbf{a}+2 \mathbf{b}$ and $3 \mathbf{a}+\lambda \mathbf{b}$ are collinear, then the set of all possible values of $\lambda$ is

Vectors $\mathbf{p}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}, \mathbf{q}=d \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $\mathbf{r}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ forming a $\triangle A B C$ are such that $\mathbf{p}=\mathbf{q}+\mathbf{r}$. If the area of $\triangle A B C$ is $5 \sqrt{6}$ sq. units, then the sum of the absolute values of $a, b, c$ is

$\mathbf{b}$ and $\mathbf{c}$ are non-collinear vectors and $(\mathbf{c} \cdot \mathbf{c}) \mathbf{a}=\mathbf{c}$. If $(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}+(\mathbf{a} \cdot \mathbf{b}) \mathbf{b}$ $=(4-2 \beta-\sin \alpha) \mathbf{b}+\left(\beta^2-1\right) \mathbf{c}$, then $\sin (\alpha+\beta)=$