If two vectors $\mathbf{a}$ and $\mathbf{b}$ which are perpendicular to each other are such that $|\mathbf{a}|=8$ and $|\mathbf{b}|=3$, then $|\mathbf{a}-2 b|=$

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