$\mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \mathbf{k}, \quad \mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ are two vectors and $\mathbf{a}$ is a vector such that $\cos (\mathbf{a}, \mathbf{b} \times \mathbf{c})=\sqrt{\frac{2}{3}}$. If $\mathbf{a}$ is a unit vector, then $|\mathbf{a} \times(\mathbf{b} \times \mathbf{c})|=$

$A(3,2,-1), B(4,1,0), C(2,1,4)$ are the vertices of a $\triangle A B C$. If the bisector of $B A C$ ! intersects the side $B C$ at $D(p, q, r)$, then $\sqrt{2 p+q+r}=$

$(3,0,2)$ and $(0,2, k)$ are the direction ratios of two lines and $\theta$ is the angle between them. If $|\cos \theta|=\frac{6}{13}$, then $k=$

$\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ 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