Given $a = i + j - k$,$b = -i + 2j + k$,and $c = -i + 2j - k$. $A$ unit vector perpendicular to both $a + b$ and $b + c$ is

Let $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ be three unit vectors such that $\vec{\alpha} \cdot \vec{\beta} = \vec{\alpha} \cdot \vec{\gamma} = 0$ and the angle between $\vec{\beta}$ and $\vec{\gamma}$ is $30^{\circ}$. Then $\vec{\alpha}$ is

The area of the triangle whose vertices are $A(1, 1, 1)$,$B(1, 2, 3)$,and $C(2, 3, 1)$ is . . . . . . .

Let $\vec{a}$ and $\vec{b}$ be two non-collinear vectors of unit modulus. If $\vec{u}=\vec{a}-(\vec{a} \cdot \vec{b}) \vec{b}$ and $\vec{v}=\vec{a} \times \vec{b}$,then $|\vec{v}|=$

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$,where $|\vec{a}|=4, |\vec{b}|=3$ and $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)$. Then $|(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})|^{2} + 4(\vec{a} \cdot \vec{b})^{2}$ is equal to

If $u = 2i + 2j - k$ and $v = 6i - 3j + 2k,$ then a unit vector perpendicular to both $u$ and $v$ is