If $a$ and $b$ are mutually perpendicular vectors,then $(a + b)^2 = $

Let $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}$ be a vector such that $\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$. Then the projection of $\vec{b}$ on the vector $\vec{a}-\vec{b}$ is :-

If $l_{1}, m_{1}, n_{1}; l_{2}, m_{2}, n_{2}; l_{3}, m_{3}, n_{3}$ are the direction cosines of three mutually perpendicular lines,prove that the line whose direction cosines are proportional to $l_{1}+l_{2}+l_{3}, m_{1}+m_{2}+m_{3}, n_{1}+n_{2}+n_{3}$ makes equal angles with them.

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 $p \times q = p \times r$ and $p \cdot q = p \cdot r$,then $\ldots . . .$.

Given $\vec{a}=3 \hat{i}-\hat{j}$,$\vec{b}=2 \hat{i}+\hat{j}-3 \hat{k}$ and $\vec{b}=\overrightarrow{b_1}+\overrightarrow{b_2}$,where $\overrightarrow{b_1}$ is parallel to $\vec{a}$ and $\overrightarrow{b_2}$ is perpendicular to $\vec{a}$,then $\overrightarrow{b_2}$ is equal to