The vectors $$\overrightarrow{\mathrm{AB}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}$$ and $$\overrightarrow{\mathrm{AC}}=5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$$ are the sides of a triangle $$\mathrm{ABC}$$. The length of the median through $$\mathrm{A}$$ is

If $$\bar{a}=2 \hat{i}+3 \hat{j}-\hat{k}, \bar{b}=-\hat{i}+2 \hat{j}-4 \hat{k}$$ and $$\bar{c}=\hat{i}+\hat{j}-2 \hat{k}$$, then $$(\bar{a} \times \bar{b}) \cdot(\bar{a} \times \bar{c})=$$

Let $$\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$$ and $$\overline{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$$. If $$\overline{\mathrm{c}}$$ is a vector such that $$\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=|\overline{\mathrm{c}}|,|\overline{\mathrm{c}}-\overline{\mathrm{a}}|=2 \sqrt{2}$$ and the angle between $$\overline{\mathrm{a}} \times \overline{\mathrm{b}}$$ and $$\overline{\mathrm{c}}$$ is $$60^{\circ}$$. Then $$|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|=$$

The projection of $$\bar{a}=\hat{i}-2 \hat{j}+\hat{k}$$ on $$\bar{b}=2 \hat{i}-\hat{j}+\hat{k}$$ is