If $a = (1, -1, 2)$,$b = (-2, 3, 5)$,$c = (2, -2, 4)$ and $i$ is the unit vector in the $x$-direction,then $(a - 2b + 3c) \cdot i = $

If $a, b, c$ are the $p^{th}, q^{th}, r^{th}$ terms of an $H.P.$ and $\vec{u} = (q-r)\hat{i} + (r-p)\hat{j} + (p-q)\hat{k}$ and $\vec{v} = \frac{\hat{i}}{a} + \frac{\hat{j}}{b} + \frac{\hat{k}}{c}$,then:

Find the angle between the following pairs of lines:
$\vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})$ and
$\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})$

If $a$ and $b$ are unlike vectors,then $a \cdot b =$

If $a, b, c$ and $r$ are vectors such that $a$ is not perpendicular to $b$,$r \times b = c \times b$ and $r \cdot a = 0$,then $r =$

Let $a=\hat{i}+2 \hat{j}-2 \hat{k}$ and $b=2 \hat{i}-\hat{j}-2 \hat{k}$. If the orthogonal projection vector of $a$ on $b$ is $x$ and the orthogonal projection vector of $b$ on $a$ is $y$, then $|x-y|=$