If $\overline{\mathrm{a}}=\frac{1}{\sqrt{10}}(4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}), \overline{\mathrm{b}}=\frac{1}{5}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$, then the value of $(2 \bar{a}-\bar{b}) \cdot\{(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b})\}$ is

If $\bar{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \bar{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k} \quad$ and $\bar{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$ are non-zero non-coplanar vectors and $m$ is non-zero scalar such that $[\mathrm{m} \overline{\mathrm{a}}+\overline{\mathrm{b}} \quad \mathrm{m} \overline{\mathrm{b}}+\overline{\mathrm{c}} \mathrm{m} \overline{\mathrm{c}}+\overline{\mathrm{a}}]=28[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]$, then the value of $m$ is equal to

If the vectors $\overline{A B}=3 \hat{i}+4 \hat{k}$ and $\overline{A C}=5 \hat{i}-2 \hat{j}+4 \hat{k}$ are the sides of the triangle $A B C$, then the length of the median through $A$ is

Let $\overline{\mathrm{a}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$. Then the vector $\overline{\mathrm{b}}$ satisfying $\overline{\mathrm{a}} \times \overline{\mathrm{b}}+\overline{\mathrm{c}}=\overline{0}$ and $\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=3$, is