If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors and the maximum value of $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$ is $k$,then $k(2|\vec{a}|^2+3|\vec{b}|^2-4|\vec{c}|^2) = $

If $\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$,then

Let $\vec{u}, \vec{v}, \vec{w}$ be vectors such that $\vec{u} + \vec{v} + \vec{w} = \vec{0}$. If $|\vec{u}| = 3$,$|\vec{v}| = 4$,and $|\vec{w}| = 5$,then $\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}$ is:

In the figure,a vector $x$ satisfies the equation $x - w = v$. Then $x =$

Let $u, v$ and $w$ be three vectors in $R^3$. Then,any vector $z \in R^3$ can be written as $z = au + bv + cw$ for some scalars $a, b$ and $c$ if and only if:

Let $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}$ and $x=2 y$. If $|\vec{a}|=5 \sqrt{2}$ and $\vec{a}$ makes an angle of $135^{\circ}$ with the $z$-axis,then $\vec{a}=$