$\mathbf{u}, \mathbf{v}$ and $\mathbf{w}$ are three unit vectors. Let $\hat{\mathbf{p}}=\hat{\mathbf{u}}+\hat{\mathbf{v}}+\hat{\mathbf{w}} \cdot \hat{\mathbf{q}}=\hat{\mathbf{u}} \times(\hat{\mathbf{v}} \times \hat{\mathbf{w}})$. If $\hat{\mathbf{p}} \cdot \hat{\mathbf{u}}=\frac{3}{2} \cdot \hat{\mathbf{p}} \hat{\mathbf{v}}=\frac{7}{4}|\hat{\mathbf{p}}|=2$ and $v=K . q$, then $K=$

If $\mathbf{a}$ and $\mathbf{b}$ are the two non collinear vectors, then $|\mathbf{b}|\mathbf{a}+|\mathbf{a}| \mathbf{b}$ represents

If $L M N$ are the mid-points of the sides $P Q, Q R$ and $R P d$ $\triangle P Q R$ respectively, then $$ \begin{aligned} & \mathbf{Q M}+\mathbf{L N}+\mathbf{M L}+\mathbf{R N}-\mathbf{M N}-\mathbf{Q L}= \end{aligned} $$

Let $\mathbf{a} \times \mathbf{b}=7 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$ and $\mathbf{a}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$. If the length of projection of $\mathbf{b}$ on $\mathbf{a}$ is $$ \frac{8}{\sqrt{14}}, \text { then }|b|= $$