Let $a = 2i + j + k$,$b = i + 2j - k$,and a unit vector $c$ be coplanar. If $c$ is perpendicular to $a$,then $c = \dots$

For any two non-zero vectors $\vec{a}$ and $\vec{b}$,$(a \vec{b} + b \vec{a}) \cdot (a \vec{b} - b \vec{a})$ is equal to:

In a trapezium $ABCD$,$\vec{BC} = \lambda \vec{AD}$ and $\vec{x} = \vec{AC} + \vec{BD}$. If $\vec{x} = p \vec{AD}$,then $p =$

In a plane, $\vec{a}$ and $\vec{b}$ are the position vectors of two points $A$ and $B$ respectively. $A$ point $P$ with position vector $\vec{r}$ moves on that plane in such a way that $|\vec{r}-\vec{a}| - |\vec{r}-\vec{b}| = c$ (where $c$ is a real constant). The locus of $P$ is a conic section whose eccentricity is:

The value of $b$ such that the scalar product of the vector $(i + j + k)$ with the unit vector parallel to the sum of the vectors $(2i + 4j - 5k)$ and $(bi + 2j + 3k)$ is $1$,is:

If $a+b+c=0$ and $|a|=5, |b|=3$ and $|c|=7$,then the angle between $a$ and $b$ is