If $a$ and $b$ are unit vectors and $\alpha$ is the angle between them, then $a+b$ is a unit vector when $\cos \alpha=$

Let $\vec a, \vec b, \vec c$ be three vectors such that $\vec a \perp (\vec b + \vec c)$,$\vec b \perp (\vec c + \vec a)$,and $\vec c \perp (\vec a + \vec b)$. If $|\vec a| = 1, |\vec b| = 2, |\vec c| = 3$,then $|\vec a + \vec b + \vec c| = \dots$

Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that $|(\hat{a}+\hat{b})+2(\hat{a} \times \hat{b})|=2$. If $\theta \in(0, \pi)$ is the angle between $\hat{a}$ and $\hat{b}$,then among the statements:
$(S_{1})$: $2|\hat{a} \times \hat{b}|=|\hat{a}-\hat{b}|$
$(S_{2})$: The projection of $\hat{a}$ on $(\hat{a}+\hat{b})$ is $\frac{1}{2}$

If $a = i + j + k$,$a \cdot b = 1$ and $a \times b = j - k$,then $b = $

If magnitudes of vectors $\vec{a}, \vec{b}, \vec{c}$ are $3, 4,$ and $5$ respectively,and $\vec{a}$ is perpendicular to $\vec{b} + \vec{c}$,$\vec{b}$ is perpendicular to $\vec{c} + \vec{a}$,and $\vec{c}$ is perpendicular to $\vec{a} + \vec{b}$,then find the value of $|\vec{a} + \vec{b} + \vec{c}|$.

If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$ such that $|\vec{a}|=7$, $|\vec{b}|=1$ and $|\vec{a} \times \vec{b}|^2 = k^2 - (\vec{a} \cdot \vec{b})^2$, then the values of $k$ and $\theta$ are