$\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ are the position vectors of the vertices $A, B$ and $C$ of a $\triangle A B C$ respectively. If $D$ and $E$ are the mid points of $B C$ and $C A$ respectively, then the unit vector along DE is

A vector of magnitude $\sqrt{2}$ units along the internal bisector of the angle between the vectors $2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ is

If $\theta$ is the angle between the vectors $4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$, then $\sin 2 \theta=$

$\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are three vectors such that $|a|=3,|b|=2 \sqrt{2},|c|=5$ and $\mathbf{c}$ is perpendicular to the plane of $\mathbf{a}$ and $\mathbf{b}$. If the angle between the vectors a and $\mathbf{b}$ is $\frac{\pi}{4}$, then $|\mathbf{a}+\mathbf{b}+\mathbf{c}|=$