Let $\vec{a} = 2\hat{i}-\hat{j}-\hat{k}$,$\vec{b} = 5\hat{i}+\hat{j}-2\hat{k}$,and $\vec{c} = -13\hat{i}-11\hat{j}+4\hat{k}$ be the position vectors of three points $A$,$B$,and $C$ respectively. If $\vec{AB} = \lambda \vec{BC}$ and $\vec{AC} = \mu \vec{CB}$,then find the value of $\lambda + \mu$.

For any two vectors $\vec{a}$ and $\vec{b},$ we always have $|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$ (triangle inequality).

If $\vec{a}$ and $\vec{b}$ are two non-collinear vectors,then $|\vec{b}| \vec{a} + |\vec{a}| \vec{b}$ represents:

The perimeter of a triangle with sides represented by vectors $\vec{a} = 3\hat{i} + 4\hat{j} + 5\hat{k}$,$\vec{b} = 4\hat{i} - 3\hat{j} - 5\hat{k}$,and $\vec{c} = 7\hat{i} + \hat{j}$ is:

The points in space represented by the position vectors $A = 4\hat{i}+\hat{j}+3\hat{k}$,$B = 6\hat{i}-2\hat{j}-3\hat{k}$,and $C = \hat{i}-\hat{j}-3\hat{k}$ form:

If $A(2 \hat{i} + \hat{j} - \hat{k})$, $B(\lambda \hat{i} + 5 \hat{j} + 4 \hat{k})$, $C(-4 \hat{i} + 3 \hat{j} + 2 \hat{k})$ and $D(-\hat{i} - 2 \hat{j} + 3 \hat{k})$ are four points in space such that $\overrightarrow{AB} = x \overrightarrow{AC} + y \overrightarrow{AD}$ for some real numbers $x \neq 0, y \neq 0$, then $17(\lambda + 9) =$ ?