$a$ and $b$ are non-collinear vectors. If $c=(x-2)a+b$ and $d=(2x+1)a-b$ are collinear vectors,then the value of $x = \ldots$

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 $\lambda + \mu$.

Let $\hat{i}-\hat{j}+2 \hat{k}$ and $\hat{i}+2 \hat{j}-2 \hat{k}$ be the position vectors of points $A$ and $B$ respectively. If $C$ is a point on the line joining $A$ and $B$ such that $BC=10$,then the position vector of $C$ can be

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors. Then the point of intersection of the line joining the points $\vec{a}+\vec{b}+\vec{c}, \vec{a}-\vec{b}+3 \vec{c}$ and the line joining the points $2 \vec{a}-\vec{b}+\vec{c}, \vec{a}-2 \vec{b}+4 \vec{c}$ is

$ABCD$ is a parallelogram such that $L$ is the mid-point of $BC$. Then,$\vec{AL}$ is equal to:

Answer the following as true or false.
Two collinear vectors are always equal in magnitude.