If $a$ is a unit vector, then $|a \times \hat{i}|^2+|a \times \hat{j}|^2+|a \times \hat{k}|^2=$

Let $\vec{r}$ be a vector in the plane of $\hat{i} - 2\hat{j} + \hat{k}$ and $\hat{i} - \hat{j} - \hat{k}$ such that $\vec{r} \cdot (\hat{i} + \hat{j}) + 2 = 0$ and the length of the projection of $\vec{r}$ on $\hat{i} - \hat{j}$ is $4\sqrt{2}$. Then,the magnitude of vector $\vec{r}$ is:

If $a, b, c$ are mutually perpendicular unit vectors,then $|a + b + c| = $

Three vectors $\vec a, \vec b, \vec c$ are inclined at an acute angle with each other such that $|\vec a| = 2, |\vec b| = 3, |\vec c| = 9$ and the lengths of the projections of $\vec a$ on $\vec b$,$\vec b$ on $\vec c$,and $\vec c$ on $\vec a$ respectively are in geometric progression. If the angle between $\vec a$ and $\vec b$ is $\frac{5\pi}{12}$ and the angle between $\vec c$ and $\vec a$ is $\frac{\pi}{12}$,then the angle between $\vec b$ and $\vec c$ is:

Given three vectors $\bar{a}, \bar{b}, \bar{c}$,two of which are collinear. If $\bar{a}+\bar{b}$ is collinear with $\bar{c}$ and $\bar{b}+\bar{c}$ is collinear with $\bar{a}$,and $|\bar{a}|=|\bar{b}|=|\bar{c}|=\sqrt{2}$,then $\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}=$

The scalar product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of the vectors $2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\lambda \hat{i}+2 \hat{j}+3 \hat{k}$ is equal to $1$. Then the value of $\lambda$ is: