$\bar{a}$ and $\bar{b}$ are non-collinear vectors. If $\bar{p} = (2x + 1)\bar{a} - \bar{b}$ and $\bar{q} = (x - 2)\bar{a} + \bar{b}$ are collinear vectors,then $x =$

Let $\vec{a}, \vec{b},$ and $\vec{c}$ be three non-zero vectors such that no two of them are collinear. If the vector $\vec{a} + 2\vec{b}$ is collinear with $\vec{c}$ and $\vec{b} + 3\vec{c}$ is collinear with $\vec{a}$,then $\vec{a} + 2\vec{b} + 6\vec{c} = \dots$

Let $u, v$ and $w$ be non-coplanar vectors. Then the points corresponding to which of the following vectors are collinear?

If $a = \hat{i} + 2 \hat{j} + 3 \hat{k}$,$b = 2 \hat{i} + 3 \hat{j} + \hat{k}$,$c = 8 \hat{i} + 13 \hat{j} + 9 \hat{k}$ and $x a + y b + z c = 0$,then $\frac{x y}{z^2} =$

Find the unit vector in the direction of vector $\overrightarrow{PQ},$ where $P$ and $Q$ are the points $(1, 2, 3)$ and $(4, 5, 6),$ respectively.

Let $u$ and $v$ be non-collinear vectors in $\mathbb{R}^2$. Let $w$ be the orthogonal projection vector of $u$ on $v$. Consider two statements:
$(i)$ Any vector in $\mathbb{R}^2$ can be written as a linear combination of $u$ and $v$.
(ii) $w$ can be written as a linear combination of $u$ and $v$ as $w = au + bv$,where both $a$ and $b$ are non-zero real numbers.