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

Let $A, B, C$ be three points whose position vectors are $\overrightarrow{a} = \hat{i} + 4\hat{j} + 3\hat{k}$,$\overrightarrow{b} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}$ (where $\alpha \in R$),and $\overrightarrow{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}$. If $\alpha$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non-collinear,then the length of the median in $\triangle ABC$ through $A$ is:

If $\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $|\bar{a}+\bar{b}|^2+|\bar{a}+\bar{c}|^2=8$,then $|\bar{a}+3\bar{b}|^2+|\bar{a}+3\bar{c}|^2=$

If $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k},$ find the unit vector in the direction of
$(i)$ $6 \vec{b}$
(ii) $2 \vec{a}-\vec{b}$

If the position vectors of points $A, B, C$ are respectively $\hat{i}, \hat{j}, \hat{k}$ and $\vec{AB} = \vec{CX}$,then the position vector of point $X$ is:

If the points with position vectors $\alpha \hat{i} + 10 \hat{j} + 13 \hat{k}$,$6 \hat{i} + 11 \hat{j} + 11 \hat{k}$,and $\frac{9}{2} \hat{i} + \beta \hat{j} - 8 \hat{k}$ are collinear,then $(19 \alpha - 6 \beta)^2$ is equal to $...........$.