$a, b, c$ are three vectors such that $|a|=1, |b|=2, |c|=3$ and $b \cdot c=0$. If the projection of $b$ along $a$ is equal to the projection of $c$ along $a$,then $|2a+3b-3c|=$

Given the following simultaneous equations for vectors $x$ and $y$:
$(i) x + y = a$
$(ii) x \times y = b$
$(iii) x \cdot a = 1$
Then $x = ?, y = ?$

Let $\vec{c}$ be the projection vector of $\vec{b}=\lambda \hat{i}+4 \hat{k}, \lambda>0$,on the vector $\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k}$. If $|\vec{a}+\vec{c}|=7$,then the area of the parallelogram formed by the vectors $\vec{b}$ and $\vec{c}$ is . . . . . . .

The projection of the vector $\hat{i} + \hat{j} + \hat{k}$ on the line $\vec{r} = 3\hat{i} - \hat{j} + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$ is:

Let a unit vector $\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$ make angles $\frac{\pi}{2}, \frac{\pi}{3}$ and $\frac{2 \pi}{3}$ with the vectors $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$ and $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$ respectively. If $\overrightarrow{v}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$,then $|\hat{u}-\overrightarrow{v}|^2$ is equal to

If $a$ and $b$ are respectively the internal and external bisectors of the angles between the vectors $u = -\hat{i} + 2\hat{j} - 2\hat{k}$ and $v = 3\hat{i} + 4\hat{j}$, and $|a| = \frac{2}{3}\sqrt{6}$, $|b| = \frac{2}{3}\sqrt{3}$, then one of the values of $a - b$ is